ABCF->ab-angle b

Percentage Accurate: 18.7% → 45.9%
Time: 26.6s
Alternatives: 13
Speedup: 5.0×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t_0} \end{array} \end{array} \]
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))
double code(double A, double B, double C, double F) {
	double t_0 = pow(B, 2.0) - ((4.0 * A) * C);
	return -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / t_0;
}
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function
public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}
def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0
function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end
function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\
\frac{-\sqrt{\left(2 \cdot \left(t_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t_0}
\end{array}
\end{array}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 13 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 18.7% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t_0} \end{array} \end{array} \]
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))
double code(double A, double B, double C, double F) {
	double t_0 = pow(B, 2.0) - ((4.0 * A) * C);
	return -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((pow((A - C), 2.0) + pow(B, 2.0)))))) / t_0;
}
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function
public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}
def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0
function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end
function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]
\begin{array}{l}

\\
\begin{array}{l}
t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\
\frac{-\sqrt{\left(2 \cdot \left(t_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t_0}
\end{array}
\end{array}

Alternative 1: 45.9% accurate, 1.4× speedup?

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := \mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)\\ \mathbf{if}\;B \leq 2.4 \cdot 10^{-101}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 3.3 \cdot 10^{-19}:\\ \;\;\;\;\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)} \cdot \frac{-1}{t_1}\\ \mathbf{elif}\;B \leq 2900000000:\\ \;\;\;\;\frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]
NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (fma B B (* C (* A -4.0)))))
   (if (<= B 2.4e-101)
     (- (/ (sqrt (* 2.0 (* (* F t_0) (* 2.0 A)))) t_0))
     (if (<= B 3.3e-19)
       (* (sqrt (* 2.0 (* t_1 (* F (- A (hypot B A)))))) (/ -1.0 t_1))
       (if (<= B 2900000000.0)
         (/
          (-
           (sqrt
            (*
             F
             (*
              (- A (fma 0.5 (/ (+ (* B B) (- (* A A) (* A A))) C) (- A)))
              (fma C (* A -8.0) (* 2.0 (* B B)))))))
          (fma B B (* A (* C -4.0))))
         (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B)))))))
B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = fma(B, B, (C * (A * -4.0)));
	double tmp;
	if (B <= 2.4e-101) {
		tmp = -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
	} else if (B <= 3.3e-19) {
		tmp = sqrt((2.0 * (t_1 * (F * (A - hypot(B, A)))))) * (-1.0 / t_1);
	} else if (B <= 2900000000.0) {
		tmp = -sqrt((F * ((A - fma(0.5, (((B * B) + ((A * A) - (A * A))) / C), -A)) * fma(C, (A * -8.0), (2.0 * (B * B)))))) / fma(B, B, (A * (C * -4.0)));
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	}
	return tmp;
}
B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_1 = fma(B, B, Float64(C * Float64(A * -4.0)))
	tmp = 0.0
	if (B <= 2.4e-101)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A)))) / t_0));
	elseif (B <= 3.3e-19)
		tmp = Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A - hypot(B, A)))))) * Float64(-1.0 / t_1));
	elseif (B <= 2900000000.0)
		tmp = Float64(Float64(-sqrt(Float64(F * Float64(Float64(A - fma(0.5, Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C), Float64(-A))) * fma(C, Float64(A * -8.0), Float64(2.0 * Float64(B * B))))))) / fma(B, B, Float64(A * Float64(C * -4.0))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end
NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(B * B + N[(C * N[(A * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.4e-101], (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), If[LessEqual[B, 3.3e-19], N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A - N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 2900000000.0], N[((-N[Sqrt[N[(F * N[(N[(A - N[(0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision] + (-A)), $MachinePrecision]), $MachinePrecision] * N[(C * N[(A * -8.0), $MachinePrecision] + N[(2.0 * N[(B * B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]]
\begin{array}{l}
B = |B|\\
[A, C] = \mathsf{sort}([A, C])\\
\\
\begin{array}{l}
t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
t_1 := \mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)\\
\mathbf{if}\;B \leq 2.4 \cdot 10^{-101}:\\
\;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\

\mathbf{elif}\;B \leq 3.3 \cdot 10^{-19}:\\
\;\;\;\;\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)} \cdot \frac{-1}{t_1}\\

\mathbf{elif}\;B \leq 2900000000:\\
\;\;\;\;\frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if B < 2.4e-101

    1. Initial program 21.3%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Step-by-step derivation
      1. Simplified21.3%

        \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
      2. Taylor expanded in A around -inf 16.8%

        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
      3. Step-by-step derivation
        1. *-commutative16.8%

          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
      4. Simplified16.8%

        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

      if 2.4e-101 < B < 3.2999999999999998e-19

      1. Initial program 51.2%

        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. Step-by-step derivation
        1. Simplified51.2%

          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
        2. Taylor expanded in C around 0 60.8%

          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        3. Step-by-step derivation
          1. +-commutative60.8%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          2. unpow260.8%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          3. unpow260.8%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          4. hypot-def60.8%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        4. Simplified60.8%

          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        5. Step-by-step derivation
          1. div-inv61.0%

            \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
          2. associate-*l*71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \color{blue}{\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          3. cancel-sign-sub-inv71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\color{blue}{\left(B \cdot B + \left(-4\right) \cdot \left(A \cdot C\right)\right)} \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          4. metadata-eval71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{-4} \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          5. associate-*r*71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{\left(-4 \cdot A\right) \cdot C}\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          6. *-commutative71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{\left(A \cdot -4\right)} \cdot C\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          7. *-commutative71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{C \cdot \left(A \cdot -4\right)}\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          8. fma-udef71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\color{blue}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          9. cancel-sign-sub-inv71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{\color{blue}{B \cdot B + \left(-4\right) \cdot \left(A \cdot C\right)}} \]
        6. Applied egg-rr71.2%

          \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}} \]
        7. Step-by-step derivation
          1. hypot-def71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \color{blue}{\sqrt{A \cdot A + B \cdot B}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
          2. unpow271.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{{A}^{2}} + B \cdot B}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
          3. unpow271.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
          4. +-commutative71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
          5. unpow271.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
          6. unpow271.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
          7. hypot-def71.2%

            \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
        8. Simplified71.2%

          \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}} \]

        if 3.2999999999999998e-19 < B < 2.9e9

        1. Initial program 1.7%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Simplified5.8%

          \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
        3. Taylor expanded in C around inf 20.8%

          \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\left(0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} + -1 \cdot A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        4. Step-by-step derivation
          1. fma-def20.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\mathsf{fma}\left(0.5, \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}, -1 \cdot A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          2. associate--l+20.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          3. unpow220.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          4. unpow220.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          5. unpow220.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          6. mul-1-neg20.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          7. mul-1-neg20.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          8. sqr-neg20.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          9. mul-1-neg20.8%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, \color{blue}{-A}\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        5. Simplified20.8%

          \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]

        if 2.9e9 < B

        1. Initial program 14.1%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Step-by-step derivation
          1. Simplified14.1%

            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
          2. Taylor expanded in C around 0 24.7%

            \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
          3. Step-by-step derivation
            1. mul-1-neg24.7%

              \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
            2. *-commutative24.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
            3. +-commutative24.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)} \]
            4. unpow224.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)} \]
            5. unpow224.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
            6. hypot-def53.8%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
          4. Simplified53.8%

            \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
        3. Recombined 4 regimes into one program.
        4. Final simplification28.7%

          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.4 \cdot 10^{-101}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 3.3 \cdot 10^{-19}:\\ \;\;\;\;\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\\ \mathbf{elif}\;B \leq 2900000000:\\ \;\;\;\;\frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

        Alternative 2: 46.5% accurate, 0.4× speedup?

        \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ t_2 := \frac{-\sqrt{\left(2 \cdot \left(t_1 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{t_1}\\ \mathbf{if}\;t_2 \leq -1 \cdot 10^{-205}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;t_2 \leq 10^{+271}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]
        NOTE: B should be positive before calling this function
        NOTE: A and C should be sorted in increasing order before calling this function.
        (FPCore (A B C F)
         :precision binary64
         (let* ((t_0 (- (* B B) (* 4.0 (* A C))))
                (t_1 (- (pow B 2.0) (* (* 4.0 A) C)))
                (t_2
                 (/
                  (-
                   (sqrt
                    (*
                     (* 2.0 (* t_1 F))
                     (- (+ A C) (sqrt (+ (pow B 2.0) (pow (- A C) 2.0)))))))
                  t_1)))
           (if (<= t_2 -1e-205)
             (/
              (*
               (sqrt (* 2.0 (* F (+ A (- C (hypot (- A C) B))))))
               (- (sqrt (fma B B (* C (* A -4.0))))))
              (fma B B (* A (* C -4.0))))
             (if (<= t_2 1e+271)
               (/
                (-
                 (sqrt
                  (*
                   2.0
                   (*
                    (* F t_0)
                    (+ A (fma -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C) A))))))
                t_0)
               (* (sqrt (* F (- C (hypot C B)))) (/ (- (sqrt 2.0)) B))))))
        B = abs(B);
        assert(A < C);
        double code(double A, double B, double C, double F) {
        	double t_0 = (B * B) - (4.0 * (A * C));
        	double t_1 = pow(B, 2.0) - ((4.0 * A) * C);
        	double t_2 = -sqrt(((2.0 * (t_1 * F)) * ((A + C) - sqrt((pow(B, 2.0) + pow((A - C), 2.0)))))) / t_1;
        	double tmp;
        	if (t_2 <= -1e-205) {
        		tmp = (sqrt((2.0 * (F * (A + (C - hypot((A - C), B)))))) * -sqrt(fma(B, B, (C * (A * -4.0))))) / fma(B, B, (A * (C * -4.0)));
        	} else if (t_2 <= 1e+271) {
        		tmp = -sqrt((2.0 * ((F * t_0) * (A + fma(-0.5, (((B * B) + ((A * A) - (A * A))) / C), A))))) / t_0;
        	} else {
        		tmp = sqrt((F * (C - hypot(C, B)))) * (-sqrt(2.0) / B);
        	}
        	return tmp;
        }
        
        B = abs(B)
        A, C = sort([A, C])
        function code(A, B, C, F)
        	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
        	t_1 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
        	t_2 = Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_1 * F)) * Float64(Float64(A + C) - sqrt(Float64((B ^ 2.0) + (Float64(A - C) ^ 2.0))))))) / t_1)
        	tmp = 0.0
        	if (t_2 <= -1e-205)
        		tmp = Float64(Float64(sqrt(Float64(2.0 * Float64(F * Float64(A + Float64(C - hypot(Float64(A - C), B)))))) * Float64(-sqrt(fma(B, B, Float64(C * Float64(A * -4.0)))))) / fma(B, B, Float64(A * Float64(C * -4.0))));
        	elseif (t_2 <= 1e+271)
        		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + fma(-0.5, Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C), A)))))) / t_0);
        	else
        		tmp = Float64(sqrt(Float64(F * Float64(C - hypot(C, B)))) * Float64(Float64(-sqrt(2.0)) / B));
        	end
        	return tmp
        end
        
        NOTE: B should be positive before calling this function
        NOTE: A and C should be sorted in increasing order before calling this function.
        code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$1 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[B, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision]}, If[LessEqual[t$95$2, -1e-205], N[(N[(N[Sqrt[N[(2.0 * N[(F * N[(A + N[(C - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[N[(B * B + N[(C * N[(A * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / N[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$2, 1e+271], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision] + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(C - N[Sqrt[C ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]]
        
        \begin{array}{l}
        B = |B|\\
        [A, C] = \mathsf{sort}([A, C])\\
        \\
        \begin{array}{l}
        t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
        t_1 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\
        t_2 := \frac{-\sqrt{\left(2 \cdot \left(t_1 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{t_1}\\
        \mathbf{if}\;t_2 \leq -1 \cdot 10^{-205}:\\
        \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\
        
        \mathbf{elif}\;t_2 \leq 10^{+271}:\\
        \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\
        
        \mathbf{else}:\\
        \;\;\;\;\sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 3 regimes
        2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C))) < -1e-205

          1. Initial program 43.9%

            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          2. Step-by-step derivation
            1. Simplified50.0%

              \[\leadsto \color{blue}{\frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)}} \]
            2. Step-by-step derivation
              1. sqrt-prod62.6%

                \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              2. associate-*r*62.6%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, \color{blue}{\left(A \cdot -4\right) \cdot C}\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              3. *-commutative62.6%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, \color{blue}{C \cdot \left(A \cdot -4\right)}\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              4. associate-*l*62.6%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{\color{blue}{2 \cdot \left(F \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              5. associate--r-63.0%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \color{blue}{\left(\left(C - \mathsf{hypot}\left(B, A - C\right)\right) + A\right)}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              6. +-commutative63.0%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \color{blue}{\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
            3. Applied egg-rr63.0%

              \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
            4. Step-by-step derivation
              1. hypot-def53.4%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \color{blue}{\sqrt{B \cdot B + \left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              2. unpow253.4%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{B \cdot B + \color{blue}{{\left(A - C\right)}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              3. unpow253.4%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{{B}^{2}} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              4. +-commutative53.4%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              5. unpow253.4%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{\left(A - C\right) \cdot \left(A - C\right)} + {B}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              6. unpow253.4%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\left(A - C\right) \cdot \left(A - C\right) + \color{blue}{B \cdot B}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
              7. hypot-def63.0%

                \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \color{blue}{\mathsf{hypot}\left(A - C, B\right)}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
            5. Simplified63.0%

              \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]

            if -1e-205 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C))) < 9.99999999999999953e270

            1. Initial program 28.2%

              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. Step-by-step derivation
              1. Simplified28.2%

                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
              2. Taylor expanded in C around inf 40.6%

                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              3. Step-by-step derivation
                1. associate--l+40.6%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} - -1 \cdot A\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                2. fma-neg40.6%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{\mathsf{fma}\left(-0.5, \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}, --1 \cdot A\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                3. associate--l+40.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                4. unpow240.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                5. unpow240.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                6. unpow240.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                7. mul-1-neg40.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                8. mul-1-neg40.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                9. sqr-neg40.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                10. mul-1-neg40.8%

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\color{blue}{\left(-A\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              4. Simplified40.8%

                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\left(-A\right)\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

              if 9.99999999999999953e270 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)))

              1. Initial program 0.3%

                \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              2. Step-by-step derivation
                1. Simplified0.3%

                  \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                2. Taylor expanded in A around 0 1.6%

                  \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F}\right)} \]
                3. Step-by-step derivation
                  1. mul-1-neg1.6%

                    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F}} \]
                  2. *-commutative1.6%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
                  3. +-commutative1.6%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{{C}^{2} + {B}^{2}}}\right)} \]
                  4. unpow21.6%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{C \cdot C} + {B}^{2}}\right)} \]
                  5. unpow21.6%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{C \cdot C + \color{blue}{B \cdot B}}\right)} \]
                  6. hypot-def17.2%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(C, B\right)}\right)} \]
                4. Simplified17.2%

                  \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)}} \]
              3. Recombined 3 regimes into one program.
              4. Final simplification37.3%

                \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq -1 \cdot 10^{-205}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq 10^{+271}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

              Alternative 3: 45.9% accurate, 1.5× speedup?

              \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ t_2 := \mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)\\ \mathbf{if}\;B \leq 2.4 \cdot 10^{-101}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 3.25 \cdot 10^{-21}:\\ \;\;\;\;\sqrt{2 \cdot \left(t_2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)} \cdot \frac{-1}{t_2}\\ \mathbf{elif}\;B \leq 1400000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]
              NOTE: B should be positive before calling this function
              NOTE: A and C should be sorted in increasing order before calling this function.
              (FPCore (A B C F)
               :precision binary64
               (let* ((t_0 (- (* B B) (* 4.0 (* A C))))
                      (t_1 (* F t_0))
                      (t_2 (fma B B (* C (* A -4.0)))))
                 (if (<= B 2.4e-101)
                   (- (/ (sqrt (* 2.0 (* t_1 (* 2.0 A)))) t_0))
                   (if (<= B 3.25e-21)
                     (* (sqrt (* 2.0 (* t_2 (* F (- A (hypot B A)))))) (/ -1.0 t_2))
                     (if (<= B 1400000000.0)
                       (/
                        (-
                         (sqrt
                          (*
                           2.0
                           (*
                            t_1
                            (+ A (fma -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C) A))))))
                        t_0)
                       (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B)))))))
              B = abs(B);
              assert(A < C);
              double code(double A, double B, double C, double F) {
              	double t_0 = (B * B) - (4.0 * (A * C));
              	double t_1 = F * t_0;
              	double t_2 = fma(B, B, (C * (A * -4.0)));
              	double tmp;
              	if (B <= 2.4e-101) {
              		tmp = -(sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
              	} else if (B <= 3.25e-21) {
              		tmp = sqrt((2.0 * (t_2 * (F * (A - hypot(B, A)))))) * (-1.0 / t_2);
              	} else if (B <= 1400000000.0) {
              		tmp = -sqrt((2.0 * (t_1 * (A + fma(-0.5, (((B * B) + ((A * A) - (A * A))) / C), A))))) / t_0;
              	} else {
              		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
              	}
              	return tmp;
              }
              
              B = abs(B)
              A, C = sort([A, C])
              function code(A, B, C, F)
              	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
              	t_1 = Float64(F * t_0)
              	t_2 = fma(B, B, Float64(C * Float64(A * -4.0)))
              	tmp = 0.0
              	if (B <= 2.4e-101)
              		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A)))) / t_0));
              	elseif (B <= 3.25e-21)
              		tmp = Float64(sqrt(Float64(2.0 * Float64(t_2 * Float64(F * Float64(A - hypot(B, A)))))) * Float64(-1.0 / t_2));
              	elseif (B <= 1400000000.0)
              		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + fma(-0.5, Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C), A)))))) / t_0);
              	else
              		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
              	end
              	return tmp
              end
              
              NOTE: B should be positive before calling this function
              NOTE: A and C should be sorted in increasing order before calling this function.
              code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(B * B + N[(C * N[(A * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.4e-101], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), If[LessEqual[B, 3.25e-21], N[(N[Sqrt[N[(2.0 * N[(t$95$2 * N[(F * N[(A - N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$2), $MachinePrecision]), $MachinePrecision], If[LessEqual[B, 1400000000.0], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision] + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]]]
              
              \begin{array}{l}
              B = |B|\\
              [A, C] = \mathsf{sort}([A, C])\\
              \\
              \begin{array}{l}
              t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
              t_1 := F \cdot t_0\\
              t_2 := \mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)\\
              \mathbf{if}\;B \leq 2.4 \cdot 10^{-101}:\\
              \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
              
              \mathbf{elif}\;B \leq 3.25 \cdot 10^{-21}:\\
              \;\;\;\;\sqrt{2 \cdot \left(t_2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)} \cdot \frac{-1}{t_2}\\
              
              \mathbf{elif}\;B \leq 1400000000:\\
              \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\
              
              \mathbf{else}:\\
              \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\
              
              
              \end{array}
              \end{array}
              
              Derivation
              1. Split input into 4 regimes
              2. if B < 2.4e-101

                1. Initial program 21.3%

                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                2. Step-by-step derivation
                  1. Simplified21.3%

                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                  2. Taylor expanded in A around -inf 16.8%

                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  3. Step-by-step derivation
                    1. *-commutative16.8%

                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  4. Simplified16.8%

                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                  if 2.4e-101 < B < 3.24999999999999993e-21

                  1. Initial program 51.2%

                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                  2. Step-by-step derivation
                    1. Simplified51.2%

                      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                    2. Taylor expanded in C around 0 60.8%

                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                    3. Step-by-step derivation
                      1. +-commutative60.8%

                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      2. unpow260.8%

                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      3. unpow260.8%

                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      4. hypot-def60.8%

                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                    4. Simplified60.8%

                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                    5. Step-by-step derivation
                      1. div-inv61.0%

                        \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                      2. associate-*l*71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \color{blue}{\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      3. cancel-sign-sub-inv71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\color{blue}{\left(B \cdot B + \left(-4\right) \cdot \left(A \cdot C\right)\right)} \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      4. metadata-eval71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{-4} \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      5. associate-*r*71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{\left(-4 \cdot A\right) \cdot C}\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      6. *-commutative71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{\left(A \cdot -4\right)} \cdot C\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      7. *-commutative71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{C \cdot \left(A \cdot -4\right)}\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      8. fma-udef71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\color{blue}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      9. cancel-sign-sub-inv71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{\color{blue}{B \cdot B + \left(-4\right) \cdot \left(A \cdot C\right)}} \]
                    6. Applied egg-rr71.2%

                      \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}} \]
                    7. Step-by-step derivation
                      1. hypot-def71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \color{blue}{\sqrt{A \cdot A + B \cdot B}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                      2. unpow271.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{{A}^{2}} + B \cdot B}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                      3. unpow271.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{{A}^{2} + \color{blue}{{B}^{2}}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                      4. +-commutative71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{{B}^{2} + {A}^{2}}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                      5. unpow271.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                      6. unpow271.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                      7. hypot-def71.2%

                        \[\leadsto \left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \]
                    8. Simplified71.2%

                      \[\leadsto \color{blue}{\left(-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)}\right) \cdot \frac{1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}} \]

                    if 3.24999999999999993e-21 < B < 1.4e9

                    1. Initial program 1.7%

                      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                    2. Step-by-step derivation
                      1. Simplified1.7%

                        \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                      2. Taylor expanded in C around inf 20.6%

                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      3. Step-by-step derivation
                        1. associate--l+20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} - -1 \cdot A\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        2. fma-neg20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{\mathsf{fma}\left(-0.5, \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}, --1 \cdot A\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        3. associate--l+20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        4. unpow220.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        5. unpow220.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        6. unpow220.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        7. mul-1-neg20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        8. mul-1-neg20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        9. sqr-neg20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                        10. mul-1-neg20.6%

                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\color{blue}{\left(-A\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                      4. Simplified20.6%

                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\left(-A\right)\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                      if 1.4e9 < B

                      1. Initial program 14.1%

                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                      2. Step-by-step derivation
                        1. Simplified14.1%

                          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                        2. Taylor expanded in C around 0 24.7%

                          \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                        3. Step-by-step derivation
                          1. mul-1-neg24.7%

                            \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                          2. *-commutative24.7%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                          3. +-commutative24.7%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)} \]
                          4. unpow224.7%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)} \]
                          5. unpow224.7%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
                          6. hypot-def53.8%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
                        4. Simplified53.8%

                          \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
                      3. Recombined 4 regimes into one program.
                      4. Final simplification28.7%

                        \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.4 \cdot 10^{-101}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 3.25 \cdot 10^{-21}:\\ \;\;\;\;\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\\ \mathbf{elif}\;B \leq 1400000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

                      Alternative 4: 46.3% accurate, 2.0× speedup?

                      \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 3.4 \cdot 10^{-29}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]
                      NOTE: B should be positive before calling this function
                      NOTE: A and C should be sorted in increasing order before calling this function.
                      (FPCore (A B C F)
                       :precision binary64
                       (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
                         (if (<= B 3.4e-29)
                           (- (/ (sqrt (* 2.0 (* (* F t_0) (* 2.0 A)))) t_0))
                           (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B)))))
                      B = abs(B);
                      assert(A < C);
                      double code(double A, double B, double C, double F) {
                      	double t_0 = (B * B) - (4.0 * (A * C));
                      	double tmp;
                      	if (B <= 3.4e-29) {
                      		tmp = -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                      	} else {
                      		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
                      	}
                      	return tmp;
                      }
                      
                      B = Math.abs(B);
                      assert A < C;
                      public static double code(double A, double B, double C, double F) {
                      	double t_0 = (B * B) - (4.0 * (A * C));
                      	double tmp;
                      	if (B <= 3.4e-29) {
                      		tmp = -(Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                      	} else {
                      		tmp = Math.sqrt((F * (A - Math.hypot(A, B)))) * (-Math.sqrt(2.0) / B);
                      	}
                      	return tmp;
                      }
                      
                      B = abs(B)
                      [A, C] = sort([A, C])
                      def code(A, B, C, F):
                      	t_0 = (B * B) - (4.0 * (A * C))
                      	tmp = 0
                      	if B <= 3.4e-29:
                      		tmp = -(math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0)
                      	else:
                      		tmp = math.sqrt((F * (A - math.hypot(A, B)))) * (-math.sqrt(2.0) / B)
                      	return tmp
                      
                      B = abs(B)
                      A, C = sort([A, C])
                      function code(A, B, C, F)
                      	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                      	tmp = 0.0
                      	if (B <= 3.4e-29)
                      		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A)))) / t_0));
                      	else
                      		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
                      	end
                      	return tmp
                      end
                      
                      B = abs(B)
                      A, C = num2cell(sort([A, C])){:}
                      function tmp_2 = code(A, B, C, F)
                      	t_0 = (B * B) - (4.0 * (A * C));
                      	tmp = 0.0;
                      	if (B <= 3.4e-29)
                      		tmp = -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                      	else
                      		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
                      	end
                      	tmp_2 = tmp;
                      end
                      
                      NOTE: B should be positive before calling this function
                      NOTE: A and C should be sorted in increasing order before calling this function.
                      code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.4e-29], (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]
                      
                      \begin{array}{l}
                      B = |B|\\
                      [A, C] = \mathsf{sort}([A, C])\\
                      \\
                      \begin{array}{l}
                      t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                      \mathbf{if}\;B \leq 3.4 \cdot 10^{-29}:\\
                      \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
                      
                      \mathbf{else}:\\
                      \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\
                      
                      
                      \end{array}
                      \end{array}
                      
                      Derivation
                      1. Split input into 2 regimes
                      2. if B < 3.39999999999999972e-29

                        1. Initial program 22.2%

                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                        2. Step-by-step derivation
                          1. Simplified22.2%

                            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                          2. Taylor expanded in A around -inf 16.8%

                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                          3. Step-by-step derivation
                            1. *-commutative16.8%

                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                          4. Simplified16.8%

                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                          if 3.39999999999999972e-29 < B

                          1. Initial program 15.4%

                            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                          2. Step-by-step derivation
                            1. Simplified15.4%

                              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                            2. Taylor expanded in C around 0 27.4%

                              \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                            3. Step-by-step derivation
                              1. mul-1-neg27.4%

                                \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                              2. *-commutative27.4%

                                \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                              3. +-commutative27.4%

                                \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)} \]
                              4. unpow227.4%

                                \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)} \]
                              5. unpow227.4%

                                \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
                              6. hypot-def53.4%

                                \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
                            4. Simplified53.4%

                              \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
                          3. Recombined 2 regimes into one program.
                          4. Final simplification27.5%

                            \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.4 \cdot 10^{-29}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

                          Alternative 5: 35.6% accurate, 2.6× speedup?

                          \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;C \leq 5 \cdot 10^{-25}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]
                          NOTE: B should be positive before calling this function
                          NOTE: A and C should be sorted in increasing order before calling this function.
                          (FPCore (A B C F)
                           :precision binary64
                           (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (+ (* B B) (* -4.0 (* A C)))))
                             (if (<= C 5e-25)
                               (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (- C (hypot B (- A C))))))))) t_1)
                               (/
                                (-
                                 (sqrt
                                  (*
                                   2.0
                                   (*
                                    (* F t_0)
                                    (+ A (fma -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C) A))))))
                                t_0))))
                          B = abs(B);
                          assert(A < C);
                          double code(double A, double B, double C, double F) {
                          	double t_0 = (B * B) - (4.0 * (A * C));
                          	double t_1 = (B * B) + (-4.0 * (A * C));
                          	double tmp;
                          	if (C <= 5e-25) {
                          		tmp = -sqrt((2.0 * (t_1 * (F * (A + (C - hypot(B, (A - C)))))))) / t_1;
                          	} else {
                          		tmp = -sqrt((2.0 * ((F * t_0) * (A + fma(-0.5, (((B * B) + ((A * A) - (A * A))) / C), A))))) / t_0;
                          	}
                          	return tmp;
                          }
                          
                          B = abs(B)
                          A, C = sort([A, C])
                          function code(A, B, C, F)
                          	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                          	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
                          	tmp = 0.0
                          	if (C <= 5e-25)
                          		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(C - hypot(B, Float64(A - C))))))))) / t_1);
                          	else
                          		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + fma(-0.5, Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C), A)))))) / t_0);
                          	end
                          	return tmp
                          end
                          
                          NOTE: B should be positive before calling this function
                          NOTE: A and C should be sorted in increasing order before calling this function.
                          code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[C, 5e-25], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(C - N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision] + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]
                          
                          \begin{array}{l}
                          B = |B|\\
                          [A, C] = \mathsf{sort}([A, C])\\
                          \\
                          \begin{array}{l}
                          t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                          t_1 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\
                          \mathbf{if}\;C \leq 5 \cdot 10^{-25}:\\
                          \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{t_1}\\
                          
                          \mathbf{else}:\\
                          \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\
                          
                          
                          \end{array}
                          \end{array}
                          
                          Derivation
                          1. Split input into 2 regimes
                          2. if C < 4.99999999999999962e-25

                            1. Initial program 27.6%

                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                            2. Step-by-step derivation
                              1. Simplified27.6%

                                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                              2. Step-by-step derivation
                                1. distribute-frac-neg27.6%

                                  \[\leadsto \color{blue}{-\frac{\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                              3. Applied egg-rr32.8%

                                \[\leadsto \color{blue}{-\frac{\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}} \]

                              if 4.99999999999999962e-25 < C

                              1. Initial program 3.4%

                                \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                              2. Step-by-step derivation
                                1. Simplified3.4%

                                  \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                2. Taylor expanded in C around inf 25.2%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                3. Step-by-step derivation
                                  1. associate--l+25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} - -1 \cdot A\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  2. fma-neg25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{\mathsf{fma}\left(-0.5, \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}, --1 \cdot A\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  3. associate--l+25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  4. unpow225.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  5. unpow225.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  6. unpow225.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  7. mul-1-neg25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  8. mul-1-neg25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  9. sqr-neg25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  10. mul-1-neg25.2%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\color{blue}{\left(-A\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                4. Simplified25.2%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\left(-A\right)\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                              3. Recombined 2 regimes into one program.
                              4. Final simplification30.5%

                                \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 5 \cdot 10^{-25}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                              Alternative 6: 32.9% accurate, 2.7× speedup?

                              \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;C \leq 3.7 \cdot 10^{+56}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \end{array} \end{array} \]
                              NOTE: B should be positive before calling this function
                              NOTE: A and C should be sorted in increasing order before calling this function.
                              (FPCore (A B C F)
                               :precision binary64
                               (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* F t_0)))
                                 (if (<= C 3.7e+56)
                                   (/ (- (sqrt (* 2.0 (* t_1 (- A (hypot A B)))))) t_0)
                                   (- (/ (sqrt (* 2.0 (* t_1 (* 2.0 A)))) t_0)))))
                              B = abs(B);
                              assert(A < C);
                              double code(double A, double B, double C, double F) {
                              	double t_0 = (B * B) - (4.0 * (A * C));
                              	double t_1 = F * t_0;
                              	double tmp;
                              	if (C <= 3.7e+56) {
                              		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
                              	} else {
                              		tmp = -(sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
                              	}
                              	return tmp;
                              }
                              
                              B = Math.abs(B);
                              assert A < C;
                              public static double code(double A, double B, double C, double F) {
                              	double t_0 = (B * B) - (4.0 * (A * C));
                              	double t_1 = F * t_0;
                              	double tmp;
                              	if (C <= 3.7e+56) {
                              		tmp = -Math.sqrt((2.0 * (t_1 * (A - Math.hypot(A, B))))) / t_0;
                              	} else {
                              		tmp = -(Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
                              	}
                              	return tmp;
                              }
                              
                              B = abs(B)
                              [A, C] = sort([A, C])
                              def code(A, B, C, F):
                              	t_0 = (B * B) - (4.0 * (A * C))
                              	t_1 = F * t_0
                              	tmp = 0
                              	if C <= 3.7e+56:
                              		tmp = -math.sqrt((2.0 * (t_1 * (A - math.hypot(A, B))))) / t_0
                              	else:
                              		tmp = -(math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0)
                              	return tmp
                              
                              B = abs(B)
                              A, C = sort([A, C])
                              function code(A, B, C, F)
                              	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                              	t_1 = Float64(F * t_0)
                              	tmp = 0.0
                              	if (C <= 3.7e+56)
                              		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A - hypot(A, B)))))) / t_0);
                              	else
                              		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A)))) / t_0));
                              	end
                              	return tmp
                              end
                              
                              B = abs(B)
                              A, C = num2cell(sort([A, C])){:}
                              function tmp_2 = code(A, B, C, F)
                              	t_0 = (B * B) - (4.0 * (A * C));
                              	t_1 = F * t_0;
                              	tmp = 0.0;
                              	if (C <= 3.7e+56)
                              		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
                              	else
                              		tmp = -(sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
                              	end
                              	tmp_2 = tmp;
                              end
                              
                              NOTE: B should be positive before calling this function
                              NOTE: A and C should be sorted in increasing order before calling this function.
                              code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, If[LessEqual[C, 3.7e+56], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision])]]]
                              
                              \begin{array}{l}
                              B = |B|\\
                              [A, C] = \mathsf{sort}([A, C])\\
                              \\
                              \begin{array}{l}
                              t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                              t_1 := F \cdot t_0\\
                              \mathbf{if}\;C \leq 3.7 \cdot 10^{+56}:\\
                              \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\
                              
                              \mathbf{else}:\\
                              \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
                              
                              
                              \end{array}
                              \end{array}
                              
                              Derivation
                              1. Split input into 2 regimes
                              2. if C < 3.69999999999999997e56

                                1. Initial program 26.3%

                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                2. Step-by-step derivation
                                  1. Simplified26.3%

                                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                  2. Taylor expanded in C around 0 22.4%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  3. Step-by-step derivation
                                    1. +-commutative22.4%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    2. unpow222.4%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    3. unpow222.4%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    4. hypot-def24.7%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  4. Simplified24.7%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                                  if 3.69999999999999997e56 < C

                                  1. Initial program 2.2%

                                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                  2. Step-by-step derivation
                                    1. Simplified2.2%

                                      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                    2. Taylor expanded in A around -inf 19.3%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    3. Step-by-step derivation
                                      1. *-commutative19.3%

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    4. Simplified19.3%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  3. Recombined 2 regimes into one program.
                                  4. Final simplification23.3%

                                    \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 3.7 \cdot 10^{+56}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                                  Alternative 7: 29.8% accurate, 2.8× speedup?

                                  \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 1.95 \cdot 10^{-28}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(B, A\right)\right) \cdot \left(F \cdot \left(B \cdot B\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]
                                  NOTE: B should be positive before calling this function
                                  NOTE: A and C should be sorted in increasing order before calling this function.
                                  (FPCore (A B C F)
                                   :precision binary64
                                   (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
                                     (if (<= B 1.95e-28)
                                       (- (/ (sqrt (* 2.0 (* (* F t_0) (* 2.0 A)))) t_0))
                                       (/ (- (sqrt (* 2.0 (* (- A (hypot B A)) (* F (* B B)))))) t_0))))
                                  B = abs(B);
                                  assert(A < C);
                                  double code(double A, double B, double C, double F) {
                                  	double t_0 = (B * B) - (4.0 * (A * C));
                                  	double tmp;
                                  	if (B <= 1.95e-28) {
                                  		tmp = -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                                  	} else {
                                  		tmp = -sqrt((2.0 * ((A - hypot(B, A)) * (F * (B * B))))) / t_0;
                                  	}
                                  	return tmp;
                                  }
                                  
                                  B = Math.abs(B);
                                  assert A < C;
                                  public static double code(double A, double B, double C, double F) {
                                  	double t_0 = (B * B) - (4.0 * (A * C));
                                  	double tmp;
                                  	if (B <= 1.95e-28) {
                                  		tmp = -(Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                                  	} else {
                                  		tmp = -Math.sqrt((2.0 * ((A - Math.hypot(B, A)) * (F * (B * B))))) / t_0;
                                  	}
                                  	return tmp;
                                  }
                                  
                                  B = abs(B)
                                  [A, C] = sort([A, C])
                                  def code(A, B, C, F):
                                  	t_0 = (B * B) - (4.0 * (A * C))
                                  	tmp = 0
                                  	if B <= 1.95e-28:
                                  		tmp = -(math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0)
                                  	else:
                                  		tmp = -math.sqrt((2.0 * ((A - math.hypot(B, A)) * (F * (B * B))))) / t_0
                                  	return tmp
                                  
                                  B = abs(B)
                                  A, C = sort([A, C])
                                  function code(A, B, C, F)
                                  	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                                  	tmp = 0.0
                                  	if (B <= 1.95e-28)
                                  		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A)))) / t_0));
                                  	else
                                  		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(A - hypot(B, A)) * Float64(F * Float64(B * B)))))) / t_0);
                                  	end
                                  	return tmp
                                  end
                                  
                                  B = abs(B)
                                  A, C = num2cell(sort([A, C])){:}
                                  function tmp_2 = code(A, B, C, F)
                                  	t_0 = (B * B) - (4.0 * (A * C));
                                  	tmp = 0.0;
                                  	if (B <= 1.95e-28)
                                  		tmp = -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                                  	else
                                  		tmp = -sqrt((2.0 * ((A - hypot(B, A)) * (F * (B * B))))) / t_0;
                                  	end
                                  	tmp_2 = tmp;
                                  end
                                  
                                  NOTE: B should be positive before calling this function
                                  NOTE: A and C should be sorted in increasing order before calling this function.
                                  code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.95e-28], (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[((-N[Sqrt[N[(2.0 * N[(N[(A - N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision] * N[(F * N[(B * B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]
                                  
                                  \begin{array}{l}
                                  B = |B|\\
                                  [A, C] = \mathsf{sort}([A, C])\\
                                  \\
                                  \begin{array}{l}
                                  t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                                  \mathbf{if}\;B \leq 1.95 \cdot 10^{-28}:\\
                                  \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
                                  
                                  \mathbf{else}:\\
                                  \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(B, A\right)\right) \cdot \left(F \cdot \left(B \cdot B\right)\right)\right)}}{t_0}\\
                                  
                                  
                                  \end{array}
                                  \end{array}
                                  
                                  Derivation
                                  1. Split input into 2 regimes
                                  2. if B < 1.94999999999999999e-28

                                    1. Initial program 22.2%

                                      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                    2. Step-by-step derivation
                                      1. Simplified22.2%

                                        \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                      2. Taylor expanded in A around -inf 16.8%

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                      3. Step-by-step derivation
                                        1. *-commutative16.8%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                      4. Simplified16.8%

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                                      if 1.94999999999999999e-28 < B

                                      1. Initial program 15.4%

                                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                      2. Step-by-step derivation
                                        1. Simplified15.4%

                                          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                        2. Taylor expanded in C around 0 15.7%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        3. Step-by-step derivation
                                          1. +-commutative15.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          2. unpow215.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          3. unpow215.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          4. hypot-def17.1%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        4. Simplified17.1%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        5. Taylor expanded in C around 0 15.8%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot \left(F \cdot {B}^{2}\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        6. Step-by-step derivation
                                          1. unpow215.8%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right) \cdot \left(F \cdot {B}^{2}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          2. unpow215.8%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right) \cdot \left(F \cdot {B}^{2}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          3. hypot-def17.4%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right) \cdot \left(F \cdot {B}^{2}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          4. unpow217.4%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(B, A\right)\right) \cdot \left(F \cdot \color{blue}{\left(B \cdot B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        7. Simplified17.4%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(\left(A - \mathsf{hypot}\left(B, A\right)\right) \cdot \left(F \cdot \left(B \cdot B\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                      3. Recombined 2 regimes into one program.
                                      4. Final simplification17.0%

                                        \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.95 \cdot 10^{-28}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(B, A\right)\right) \cdot \left(F \cdot \left(B \cdot B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                                      Alternative 8: 28.3% accurate, 4.8× speedup?

                                      \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;B \leq 2100000000:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\ \end{array} \end{array} \]
                                      NOTE: B should be positive before calling this function
                                      NOTE: A and C should be sorted in increasing order before calling this function.
                                      (FPCore (A B C F)
                                       :precision binary64
                                       (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* F t_0)))
                                         (if (<= B 2100000000.0)
                                           (- (/ (sqrt (* 2.0 (* t_1 (* 2.0 A)))) t_0))
                                           (/ (- (sqrt (* 2.0 (* t_1 (- (+ A C) B))))) t_0))))
                                      B = abs(B);
                                      assert(A < C);
                                      double code(double A, double B, double C, double F) {
                                      	double t_0 = (B * B) - (4.0 * (A * C));
                                      	double t_1 = F * t_0;
                                      	double tmp;
                                      	if (B <= 2100000000.0) {
                                      		tmp = -(sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
                                      	} else {
                                      		tmp = -sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0;
                                      	}
                                      	return tmp;
                                      }
                                      
                                      NOTE: B should be positive before calling this function
                                      NOTE: A and C should be sorted in increasing order before calling this function.
                                      real(8) function code(a, b, c, f)
                                          real(8), intent (in) :: a
                                          real(8), intent (in) :: b
                                          real(8), intent (in) :: c
                                          real(8), intent (in) :: f
                                          real(8) :: t_0
                                          real(8) :: t_1
                                          real(8) :: tmp
                                          t_0 = (b * b) - (4.0d0 * (a * c))
                                          t_1 = f * t_0
                                          if (b <= 2100000000.0d0) then
                                              tmp = -(sqrt((2.0d0 * (t_1 * (2.0d0 * a)))) / t_0)
                                          else
                                              tmp = -sqrt((2.0d0 * (t_1 * ((a + c) - b)))) / t_0
                                          end if
                                          code = tmp
                                      end function
                                      
                                      B = Math.abs(B);
                                      assert A < C;
                                      public static double code(double A, double B, double C, double F) {
                                      	double t_0 = (B * B) - (4.0 * (A * C));
                                      	double t_1 = F * t_0;
                                      	double tmp;
                                      	if (B <= 2100000000.0) {
                                      		tmp = -(Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
                                      	} else {
                                      		tmp = -Math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0;
                                      	}
                                      	return tmp;
                                      }
                                      
                                      B = abs(B)
                                      [A, C] = sort([A, C])
                                      def code(A, B, C, F):
                                      	t_0 = (B * B) - (4.0 * (A * C))
                                      	t_1 = F * t_0
                                      	tmp = 0
                                      	if B <= 2100000000.0:
                                      		tmp = -(math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0)
                                      	else:
                                      		tmp = -math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0
                                      	return tmp
                                      
                                      B = abs(B)
                                      A, C = sort([A, C])
                                      function code(A, B, C, F)
                                      	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                                      	t_1 = Float64(F * t_0)
                                      	tmp = 0.0
                                      	if (B <= 2100000000.0)
                                      		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A)))) / t_0));
                                      	else
                                      		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(Float64(A + C) - B))))) / t_0);
                                      	end
                                      	return tmp
                                      end
                                      
                                      B = abs(B)
                                      A, C = num2cell(sort([A, C])){:}
                                      function tmp_2 = code(A, B, C, F)
                                      	t_0 = (B * B) - (4.0 * (A * C));
                                      	t_1 = F * t_0;
                                      	tmp = 0.0;
                                      	if (B <= 2100000000.0)
                                      		tmp = -(sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0);
                                      	else
                                      		tmp = -sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0;
                                      	end
                                      	tmp_2 = tmp;
                                      end
                                      
                                      NOTE: B should be positive before calling this function
                                      NOTE: A and C should be sorted in increasing order before calling this function.
                                      code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, If[LessEqual[B, 2100000000.0], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(N[(A + C), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]
                                      
                                      \begin{array}{l}
                                      B = |B|\\
                                      [A, C] = \mathsf{sort}([A, C])\\
                                      \\
                                      \begin{array}{l}
                                      t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                                      t_1 := F \cdot t_0\\
                                      \mathbf{if}\;B \leq 2100000000:\\
                                      \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
                                      
                                      \mathbf{else}:\\
                                      \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\
                                      
                                      
                                      \end{array}
                                      \end{array}
                                      
                                      Derivation
                                      1. Split input into 2 regimes
                                      2. if B < 2.1e9

                                        1. Initial program 22.4%

                                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                        2. Step-by-step derivation
                                          1. Simplified22.4%

                                            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                          2. Taylor expanded in A around -inf 16.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          3. Step-by-step derivation
                                            1. *-commutative16.7%

                                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          4. Simplified16.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                                          if 2.1e9 < B

                                          1. Initial program 14.1%

                                            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                          2. Step-by-step derivation
                                            1. Simplified14.1%

                                              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                            2. Taylor expanded in B around inf 13.2%

                                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \color{blue}{B}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          3. Recombined 2 regimes into one program.
                                          4. Final simplification15.8%

                                            \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2100000000:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                                          Alternative 9: 26.7% accurate, 5.0× speedup?

                                          \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ -\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0} \end{array} \end{array} \]
                                          NOTE: B should be positive before calling this function
                                          NOTE: A and C should be sorted in increasing order before calling this function.
                                          (FPCore (A B C F)
                                           :precision binary64
                                           (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
                                             (- (/ (sqrt (* 2.0 (* (* F t_0) (* 2.0 A)))) t_0))))
                                          B = abs(B);
                                          assert(A < C);
                                          double code(double A, double B, double C, double F) {
                                          	double t_0 = (B * B) - (4.0 * (A * C));
                                          	return -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                                          }
                                          
                                          NOTE: B should be positive before calling this function
                                          NOTE: A and C should be sorted in increasing order before calling this function.
                                          real(8) function code(a, b, c, f)
                                              real(8), intent (in) :: a
                                              real(8), intent (in) :: b
                                              real(8), intent (in) :: c
                                              real(8), intent (in) :: f
                                              real(8) :: t_0
                                              t_0 = (b * b) - (4.0d0 * (a * c))
                                              code = -(sqrt((2.0d0 * ((f * t_0) * (2.0d0 * a)))) / t_0)
                                          end function
                                          
                                          B = Math.abs(B);
                                          assert A < C;
                                          public static double code(double A, double B, double C, double F) {
                                          	double t_0 = (B * B) - (4.0 * (A * C));
                                          	return -(Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                                          }
                                          
                                          B = abs(B)
                                          [A, C] = sort([A, C])
                                          def code(A, B, C, F):
                                          	t_0 = (B * B) - (4.0 * (A * C))
                                          	return -(math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0)
                                          
                                          B = abs(B)
                                          A, C = sort([A, C])
                                          function code(A, B, C, F)
                                          	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                                          	return Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A)))) / t_0))
                                          end
                                          
                                          B = abs(B)
                                          A, C = num2cell(sort([A, C])){:}
                                          function tmp = code(A, B, C, F)
                                          	t_0 = (B * B) - (4.0 * (A * C));
                                          	tmp = -(sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0);
                                          end
                                          
                                          NOTE: B should be positive before calling this function
                                          NOTE: A and C should be sorted in increasing order before calling this function.
                                          code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision])]
                                          
                                          \begin{array}{l}
                                          B = |B|\\
                                          [A, C] = \mathsf{sort}([A, C])\\
                                          \\
                                          \begin{array}{l}
                                          t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                                          -\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}
                                          \end{array}
                                          \end{array}
                                          
                                          Derivation
                                          1. Initial program 20.2%

                                            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                          2. Step-by-step derivation
                                            1. Simplified20.2%

                                              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                            2. Taylor expanded in A around -inf 12.9%

                                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                            3. Step-by-step derivation
                                              1. *-commutative12.9%

                                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                            4. Simplified12.9%

                                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                            5. Final simplification12.9%

                                              \[\leadsto -\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                                            Alternative 10: 2.1% accurate, 5.2× speedup?

                                            \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot \left(C \cdot C\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \end{array} \]
                                            NOTE: B should be positive before calling this function
                                            NOTE: A and C should be sorted in increasing order before calling this function.
                                            (FPCore (A B C F)
                                             :precision binary64
                                             (/
                                              (- (sqrt (* 2.0 (* -8.0 (* A (* F (* C C)))))))
                                              (- (* B B) (* 4.0 (* A C)))))
                                            B = abs(B);
                                            assert(A < C);
                                            double code(double A, double B, double C, double F) {
                                            	return -sqrt((2.0 * (-8.0 * (A * (F * (C * C)))))) / ((B * B) - (4.0 * (A * C)));
                                            }
                                            
                                            NOTE: B should be positive before calling this function
                                            NOTE: A and C should be sorted in increasing order before calling this function.
                                            real(8) function code(a, b, c, f)
                                                real(8), intent (in) :: a
                                                real(8), intent (in) :: b
                                                real(8), intent (in) :: c
                                                real(8), intent (in) :: f
                                                code = -sqrt((2.0d0 * ((-8.0d0) * (a * (f * (c * c)))))) / ((b * b) - (4.0d0 * (a * c)))
                                            end function
                                            
                                            B = Math.abs(B);
                                            assert A < C;
                                            public static double code(double A, double B, double C, double F) {
                                            	return -Math.sqrt((2.0 * (-8.0 * (A * (F * (C * C)))))) / ((B * B) - (4.0 * (A * C)));
                                            }
                                            
                                            B = abs(B)
                                            [A, C] = sort([A, C])
                                            def code(A, B, C, F):
                                            	return -math.sqrt((2.0 * (-8.0 * (A * (F * (C * C)))))) / ((B * B) - (4.0 * (A * C)))
                                            
                                            B = abs(B)
                                            A, C = sort([A, C])
                                            function code(A, B, C, F)
                                            	return Float64(Float64(-sqrt(Float64(2.0 * Float64(-8.0 * Float64(A * Float64(F * Float64(C * C))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))))
                                            end
                                            
                                            B = abs(B)
                                            A, C = num2cell(sort([A, C])){:}
                                            function tmp = code(A, B, C, F)
                                            	tmp = -sqrt((2.0 * (-8.0 * (A * (F * (C * C)))))) / ((B * B) - (4.0 * (A * C)));
                                            end
                                            
                                            NOTE: B should be positive before calling this function
                                            NOTE: A and C should be sorted in increasing order before calling this function.
                                            code[A_, B_, C_, F_] := N[((-N[Sqrt[N[(2.0 * N[(-8.0 * N[(A * N[(F * N[(C * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
                                            
                                            \begin{array}{l}
                                            B = |B|\\
                                            [A, C] = \mathsf{sort}([A, C])\\
                                            \\
                                            \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot \left(C \cdot C\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}
                                            \end{array}
                                            
                                            Derivation
                                            1. Initial program 20.2%

                                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                            2. Step-by-step derivation
                                              1. Simplified20.2%

                                                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                              2. Taylor expanded in C around -inf 10.1%

                                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                              3. Step-by-step derivation
                                                1. *-commutative10.1%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                              4. Simplified10.1%

                                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                              5. Taylor expanded in B around 0 7.2%

                                                \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(-8 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                              6. Step-by-step derivation
                                                1. unpow27.2%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(A \cdot \left(\color{blue}{\left(C \cdot C\right)} \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                              7. Simplified7.2%

                                                \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(-8 \cdot \left(A \cdot \left(\left(C \cdot C\right) \cdot F\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                              8. Final simplification7.2%

                                                \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(A \cdot \left(F \cdot \left(C \cdot C\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                                              Alternative 11: 15.9% accurate, 5.2× speedup?

                                              \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \end{array} \]
                                              NOTE: B should be positive before calling this function
                                              NOTE: A and C should be sorted in increasing order before calling this function.
                                              (FPCore (A B C F)
                                               :precision binary64
                                               (/
                                                (- (sqrt (* 2.0 (* -8.0 (* (* A A) (* C F))))))
                                                (- (* B B) (* 4.0 (* A C)))))
                                              B = abs(B);
                                              assert(A < C);
                                              double code(double A, double B, double C, double F) {
                                              	return -sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)));
                                              }
                                              
                                              NOTE: B should be positive before calling this function
                                              NOTE: A and C should be sorted in increasing order before calling this function.
                                              real(8) function code(a, b, c, f)
                                                  real(8), intent (in) :: a
                                                  real(8), intent (in) :: b
                                                  real(8), intent (in) :: c
                                                  real(8), intent (in) :: f
                                                  code = -sqrt((2.0d0 * ((-8.0d0) * ((a * a) * (c * f))))) / ((b * b) - (4.0d0 * (a * c)))
                                              end function
                                              
                                              B = Math.abs(B);
                                              assert A < C;
                                              public static double code(double A, double B, double C, double F) {
                                              	return -Math.sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)));
                                              }
                                              
                                              B = abs(B)
                                              [A, C] = sort([A, C])
                                              def code(A, B, C, F):
                                              	return -math.sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)))
                                              
                                              B = abs(B)
                                              A, C = sort([A, C])
                                              function code(A, B, C, F)
                                              	return Float64(Float64(-sqrt(Float64(2.0 * Float64(-8.0 * Float64(Float64(A * A) * Float64(C * F)))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))))
                                              end
                                              
                                              B = abs(B)
                                              A, C = num2cell(sort([A, C])){:}
                                              function tmp = code(A, B, C, F)
                                              	tmp = -sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)));
                                              end
                                              
                                              NOTE: B should be positive before calling this function
                                              NOTE: A and C should be sorted in increasing order before calling this function.
                                              code[A_, B_, C_, F_] := N[((-N[Sqrt[N[(2.0 * N[(-8.0 * N[(N[(A * A), $MachinePrecision] * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
                                              
                                              \begin{array}{l}
                                              B = |B|\\
                                              [A, C] = \mathsf{sort}([A, C])\\
                                              \\
                                              \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}
                                              \end{array}
                                              
                                              Derivation
                                              1. Initial program 20.2%

                                                \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                              2. Step-by-step derivation
                                                1. Simplified20.2%

                                                  \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                                2. Taylor expanded in C around 0 17.7%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                3. Step-by-step derivation
                                                  1. +-commutative17.7%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                  2. unpow217.7%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                  3. unpow217.7%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                  4. hypot-def19.6%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                4. Simplified19.6%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                5. Taylor expanded in A around -inf 8.7%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(-8 \cdot \left({A}^{2} \cdot \left(C \cdot F\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                6. Step-by-step derivation
                                                  1. unpow28.7%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\color{blue}{\left(A \cdot A\right)} \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                7. Simplified8.7%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                8. Final simplification8.7%

                                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]

                                                Alternative 12: 1.3% accurate, 5.8× speedup?

                                                \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -2 \cdot \left({\left(C \cdot F\right)}^{0.5} \cdot \frac{1}{B}\right) \end{array} \]
                                                NOTE: B should be positive before calling this function
                                                NOTE: A and C should be sorted in increasing order before calling this function.
                                                (FPCore (A B C F) :precision binary64 (* -2.0 (* (pow (* C F) 0.5) (/ 1.0 B))))
                                                B = abs(B);
                                                assert(A < C);
                                                double code(double A, double B, double C, double F) {
                                                	return -2.0 * (pow((C * F), 0.5) * (1.0 / B));
                                                }
                                                
                                                NOTE: B should be positive before calling this function
                                                NOTE: A and C should be sorted in increasing order before calling this function.
                                                real(8) function code(a, b, c, f)
                                                    real(8), intent (in) :: a
                                                    real(8), intent (in) :: b
                                                    real(8), intent (in) :: c
                                                    real(8), intent (in) :: f
                                                    code = (-2.0d0) * (((c * f) ** 0.5d0) * (1.0d0 / b))
                                                end function
                                                
                                                B = Math.abs(B);
                                                assert A < C;
                                                public static double code(double A, double B, double C, double F) {
                                                	return -2.0 * (Math.pow((C * F), 0.5) * (1.0 / B));
                                                }
                                                
                                                B = abs(B)
                                                [A, C] = sort([A, C])
                                                def code(A, B, C, F):
                                                	return -2.0 * (math.pow((C * F), 0.5) * (1.0 / B))
                                                
                                                B = abs(B)
                                                A, C = sort([A, C])
                                                function code(A, B, C, F)
                                                	return Float64(-2.0 * Float64((Float64(C * F) ^ 0.5) * Float64(1.0 / B)))
                                                end
                                                
                                                B = abs(B)
                                                A, C = num2cell(sort([A, C])){:}
                                                function tmp = code(A, B, C, F)
                                                	tmp = -2.0 * (((C * F) ^ 0.5) * (1.0 / B));
                                                end
                                                
                                                NOTE: B should be positive before calling this function
                                                NOTE: A and C should be sorted in increasing order before calling this function.
                                                code[A_, B_, C_, F_] := N[(-2.0 * N[(N[Power[N[(C * F), $MachinePrecision], 0.5], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
                                                
                                                \begin{array}{l}
                                                B = |B|\\
                                                [A, C] = \mathsf{sort}([A, C])\\
                                                \\
                                                -2 \cdot \left({\left(C \cdot F\right)}^{0.5} \cdot \frac{1}{B}\right)
                                                \end{array}
                                                
                                                Derivation
                                                1. Initial program 20.2%

                                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                                2. Step-by-step derivation
                                                  1. Simplified20.2%

                                                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                                  2. Taylor expanded in C around -inf 10.1%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                  3. Step-by-step derivation
                                                    1. *-commutative10.1%

                                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                  4. Simplified10.1%

                                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                  5. Taylor expanded in B around inf 2.9%

                                                    \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{C \cdot F} \cdot \frac{1}{B}\right)} \]
                                                  6. Step-by-step derivation
                                                    1. pow1/23.1%

                                                      \[\leadsto -2 \cdot \left(\color{blue}{{\left(C \cdot F\right)}^{0.5}} \cdot \frac{1}{B}\right) \]
                                                    2. *-commutative3.1%

                                                      \[\leadsto -2 \cdot \left({\color{blue}{\left(F \cdot C\right)}}^{0.5} \cdot \frac{1}{B}\right) \]
                                                  7. Applied egg-rr3.1%

                                                    \[\leadsto -2 \cdot \left(\color{blue}{{\left(F \cdot C\right)}^{0.5}} \cdot \frac{1}{B}\right) \]
                                                  8. Final simplification3.1%

                                                    \[\leadsto -2 \cdot \left({\left(C \cdot F\right)}^{0.5} \cdot \frac{1}{B}\right) \]

                                                  Alternative 13: 1.0% accurate, 5.9× speedup?

                                                  \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -2 \cdot \frac{\sqrt{C \cdot F}}{B} \end{array} \]
                                                  NOTE: B should be positive before calling this function
                                                  NOTE: A and C should be sorted in increasing order before calling this function.
                                                  (FPCore (A B C F) :precision binary64 (* -2.0 (/ (sqrt (* C F)) B)))
                                                  B = abs(B);
                                                  assert(A < C);
                                                  double code(double A, double B, double C, double F) {
                                                  	return -2.0 * (sqrt((C * F)) / B);
                                                  }
                                                  
                                                  NOTE: B should be positive before calling this function
                                                  NOTE: A and C should be sorted in increasing order before calling this function.
                                                  real(8) function code(a, b, c, f)
                                                      real(8), intent (in) :: a
                                                      real(8), intent (in) :: b
                                                      real(8), intent (in) :: c
                                                      real(8), intent (in) :: f
                                                      code = (-2.0d0) * (sqrt((c * f)) / b)
                                                  end function
                                                  
                                                  B = Math.abs(B);
                                                  assert A < C;
                                                  public static double code(double A, double B, double C, double F) {
                                                  	return -2.0 * (Math.sqrt((C * F)) / B);
                                                  }
                                                  
                                                  B = abs(B)
                                                  [A, C] = sort([A, C])
                                                  def code(A, B, C, F):
                                                  	return -2.0 * (math.sqrt((C * F)) / B)
                                                  
                                                  B = abs(B)
                                                  A, C = sort([A, C])
                                                  function code(A, B, C, F)
                                                  	return Float64(-2.0 * Float64(sqrt(Float64(C * F)) / B))
                                                  end
                                                  
                                                  B = abs(B)
                                                  A, C = num2cell(sort([A, C])){:}
                                                  function tmp = code(A, B, C, F)
                                                  	tmp = -2.0 * (sqrt((C * F)) / B);
                                                  end
                                                  
                                                  NOTE: B should be positive before calling this function
                                                  NOTE: A and C should be sorted in increasing order before calling this function.
                                                  code[A_, B_, C_, F_] := N[(-2.0 * N[(N[Sqrt[N[(C * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]
                                                  
                                                  \begin{array}{l}
                                                  B = |B|\\
                                                  [A, C] = \mathsf{sort}([A, C])\\
                                                  \\
                                                  -2 \cdot \frac{\sqrt{C \cdot F}}{B}
                                                  \end{array}
                                                  
                                                  Derivation
                                                  1. Initial program 20.2%

                                                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                                  2. Step-by-step derivation
                                                    1. Simplified20.2%

                                                      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                                    2. Taylor expanded in C around -inf 10.1%

                                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot C\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                    3. Step-by-step derivation
                                                      1. *-commutative10.1%

                                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                    4. Simplified10.1%

                                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(C \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                                    5. Taylor expanded in B around inf 2.9%

                                                      \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{C \cdot F} \cdot \frac{1}{B}\right)} \]
                                                    6. Step-by-step derivation
                                                      1. un-div-inv3.0%

                                                        \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{C \cdot F}}{B}} \]
                                                      2. *-commutative3.0%

                                                        \[\leadsto -2 \cdot \frac{\sqrt{\color{blue}{F \cdot C}}}{B} \]
                                                    7. Applied egg-rr3.0%

                                                      \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{F \cdot C}}{B}} \]
                                                    8. Final simplification3.0%

                                                      \[\leadsto -2 \cdot \frac{\sqrt{C \cdot F}}{B} \]

                                                    Reproduce

                                                    ?
                                                    herbie shell --seed 2023252 
                                                    (FPCore (A B C F)
                                                      :name "ABCF->ab-angle b"
                                                      :precision binary64
                                                      (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))