ABCF->ab-angle b

Percentage Accurate: 19.4% → 50.6%
Time: 20.9s
Alternatives: 10
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 10 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: 19.4% 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: 50.6% accurate, 0.3× speedup?

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\\ t_2 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ t_3 := \frac{-\sqrt{\left(2 \cdot \left(F \cdot t_2\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{t_2}\\ \mathbf{if}\;t_3 \leq -2 \cdot 10^{-205}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{t_1}\\ \mathbf{elif}\;t_3 \leq 10^{+201}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\ \mathbf{elif}\;t_3 \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, B \cdot \left(2 \cdot B\right)\right)} \cdot \left(-\sqrt{F}\right)}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)\\ \end{array} \end{array} \]
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 (* A (* C -4.0))))
        (t_2 (- (pow B 2.0) (* (* 4.0 A) C)))
        (t_3
         (/
          (-
           (sqrt
            (*
             (* 2.0 (* F t_2))
             (- (+ A C) (sqrt (+ (pow B 2.0) (pow (- A C) 2.0)))))))
          t_2)))
   (if (<= t_3 -2e-205)
     (/
      (*
       (sqrt (* 2.0 (* F (+ A (- C (hypot (- A C) B))))))
       (- (sqrt (fma B B (* C (* A -4.0))))))
      t_1)
     (if (<= t_3 1e+201)
       (/
        (-
         (sqrt
          (*
           2.0
           (*
            (* F t_0)
            (+ A (fma -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C) A))))))
        t_0)
       (if (<= t_3 INFINITY)
         (/
          (*
           (sqrt
            (*
             (+ C (- A (hypot B (- A C))))
             (fma C (* A -8.0) (* B (* 2.0 B)))))
           (- (sqrt F)))
          t_1)
         (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A (hypot A 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, (A * (C * -4.0)));
	double t_2 = pow(B, 2.0) - ((4.0 * A) * C);
	double t_3 = -sqrt(((2.0 * (F * t_2)) * ((A + C) - sqrt((pow(B, 2.0) + pow((A - C), 2.0)))))) / t_2;
	double tmp;
	if (t_3 <= -2e-205) {
		tmp = (sqrt((2.0 * (F * (A + (C - hypot((A - C), B)))))) * -sqrt(fma(B, B, (C * (A * -4.0))))) / t_1;
	} else if (t_3 <= 1e+201) {
		tmp = -sqrt((2.0 * ((F * t_0) * (A + fma(-0.5, (((B * B) + ((A * A) - (A * A))) / C), A))))) / t_0;
	} else if (t_3 <= ((double) INFINITY)) {
		tmp = (sqrt(((C + (A - hypot(B, (A - C)))) * fma(C, (A * -8.0), (B * (2.0 * B))))) * -sqrt(F)) / t_1;
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - hypot(A, 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(A * Float64(C * -4.0)))
	t_2 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	t_3 = Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(F * t_2)) * Float64(Float64(A + C) - sqrt(Float64((B ^ 2.0) + (Float64(A - C) ^ 2.0))))))) / t_2)
	tmp = 0.0
	if (t_3 <= -2e-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)))))) / t_1);
	elseif (t_3 <= 1e+201)
		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);
	elseif (t_3 <= Inf)
		tmp = Float64(Float64(sqrt(Float64(Float64(C + Float64(A - hypot(B, Float64(A - C)))) * fma(C, Float64(A * -8.0), Float64(B * Float64(2.0 * B))))) * Float64(-sqrt(F))) / t_1);
	else
		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(F * Float64(A - hypot(A, 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[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[((-N[Sqrt[N[(N[(2.0 * N[(F * t$95$2), $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$2), $MachinePrecision]}, If[LessEqual[t$95$3, -2e-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] / t$95$1), $MachinePrecision], If[LessEqual[t$95$3, 1e+201], 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], If[LessEqual[t$95$3, Infinity], N[(N[(N[Sqrt[N[(N[(C + N[(A - N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(C * N[(A * -8.0), $MachinePrecision] + N[(B * N[(2.0 * B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[F], $MachinePrecision])), $MachinePrecision] / t$95$1), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $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, A \cdot \left(C \cdot -4\right)\right)\\
t_2 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\
t_3 := \frac{-\sqrt{\left(2 \cdot \left(F \cdot t_2\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{t_2}\\
\mathbf{if}\;t_3 \leq -2 \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)}{t_1}\\

\mathbf{elif}\;t_3 \leq 10^{+201}:\\
\;\;\;\;\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{elif}\;t_3 \leq \infty:\\
\;\;\;\;\frac{\sqrt{\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, B \cdot \left(2 \cdot B\right)\right)} \cdot \left(-\sqrt{F}\right)}{t_1}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 4 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))) < -2e-205

    1. Initial program 42.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Step-by-step derivation
      1. Simplified51.5%

        \[\leadsto \color{blue}{\frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)}} \]
      2. Step-by-step derivation
        1. sqrt-prod67.5%

          \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        2. associate-*r*67.5%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, \color{blue}{\left(A \cdot -4\right) \cdot C}\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        3. *-commutative67.5%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, \color{blue}{C \cdot \left(A \cdot -4\right)}\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        4. associate-*l*67.5%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{\color{blue}{2 \cdot \left(F \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        5. associate--r-67.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \color{blue}{\left(\left(C - \mathsf{hypot}\left(B, A - C\right)\right) + A\right)}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        6. +-commutative67.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \color{blue}{\left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
      3. Applied egg-rr67.2%

        \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
      4. Step-by-step derivation
        1. hypot-def54.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \color{blue}{\sqrt{B \cdot B + \left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        2. unpow254.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{B \cdot B + \color{blue}{{\left(A - C\right)}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        3. unpow254.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{{B}^{2}} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        4. +-commutative54.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        5. unpow254.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{\left(A - C\right) \cdot \left(A - C\right)} + {B}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        6. unpow254.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\left(A - C\right) \cdot \left(A - C\right) + \color{blue}{B \cdot B}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        7. hypot-def67.2%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \color{blue}{\mathsf{hypot}\left(A - C, B\right)}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
      5. Simplified67.2%

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

      if -2e-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))) < 1.00000000000000004e201

      1. Initial program 28.4%

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

          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
        2. Taylor expanded in C around inf 43.9%

          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        3. Step-by-step derivation
          1. associate--l+43.9%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} - -1 \cdot A\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          2. fma-neg43.9%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \color{blue}{\mathsf{fma}\left(-0.5, \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}, --1 \cdot A\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          3. associate--l+44.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          4. unpow244.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          5. unpow244.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          6. unpow244.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          7. mul-1-neg44.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          8. mul-1-neg44.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          9. sqr-neg44.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C}, --1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          10. mul-1-neg44.2%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, -\color{blue}{\left(-A\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        4. Simplified44.2%

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

        if 1.00000000000000004e201 < (/.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))) < +inf.0

        1. Initial program 4.4%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Simplified22.3%

          \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
        3. Step-by-step derivation
          1. sqrt-prod57.8%

            \[\leadsto \frac{-\color{blue}{\sqrt{F} \cdot \sqrt{\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          2. associate--r-57.8%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\color{blue}{\left(\left(A - \mathsf{hypot}\left(B, A - C\right)\right) + C\right)} \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          3. associate-*r*57.8%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\left(\left(A - \mathsf{hypot}\left(B, A - C\right)\right) + C\right) \cdot \mathsf{fma}\left(C, A \cdot -8, \color{blue}{\left(2 \cdot B\right) \cdot B}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        4. Applied egg-rr57.8%

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

        if +inf.0 < (/.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 0.0%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Step-by-step derivation
          1. Simplified0.0%

            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
          2. Taylor expanded in C around 0 1.7%

            \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
          3. Step-by-step derivation
            1. mul-1-neg1.7%

              \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
            2. *-commutative1.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
            3. +-commutative1.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{{A}^{2} + {B}^{2}}}\right)} \]
            4. unpow21.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)} \]
            5. unpow21.7%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
            6. hypot-def18.1%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
          4. Simplified18.1%

            \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
        3. Recombined 4 regimes into one program.
        4. Final simplification42.1%

          \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{-\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq -2 \cdot 10^{-205}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq 10^{+201}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(F \cdot \left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(C + \left(A - \mathsf{hypot}\left(B, A - C\right)\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, B \cdot \left(2 \cdot B\right)\right)} \cdot \left(-\sqrt{F}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)\\ \end{array} \]

        Alternative 2: 47.4% accurate, 2.0× speedup?

        \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 2.7 \cdot 10^{-31}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(2 \cdot \left(A \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]
        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
         (if (<= B 2.7e-31)
           (/
            (- (sqrt (* 2.0 (* (fma B B (* A (* C -4.0))) (* 2.0 (* A F))))))
            (- (* B B) (* 4.0 (* A C))))
           (* (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 tmp;
        	if (B <= 2.7e-31) {
        		tmp = -sqrt((2.0 * (fma(B, B, (A * (C * -4.0))) * (2.0 * (A * F))))) / ((B * B) - (4.0 * (A * C)));
        	} 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)
        	tmp = 0.0
        	if (B <= 2.7e-31)
        		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(fma(B, B, Float64(A * Float64(C * -4.0))) * Float64(2.0 * Float64(A * F)))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
        	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_] := If[LessEqual[B, 2.7e-31], N[((-N[Sqrt[N[(2.0 * N[(N[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(A * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $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}
        \mathbf{if}\;B \leq 2.7 \cdot 10^{-31}:\\
        \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(2 \cdot \left(A \cdot F\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}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if B < 2.70000000000000014e-31

          1. Initial program 20.3%

            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          2. Step-by-step derivation
            1. Simplified20.3%

              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
            2. Taylor expanded in A around -inf 16.4%

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

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

              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
            5. Step-by-step derivation
              1. *-un-lft-identity16.4%

                \[\leadsto \frac{-\color{blue}{1 \cdot \sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A \cdot 2\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              2. associate-*l*17.4%

                \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \color{blue}{\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              3. cancel-sign-sub-inv17.4%

                \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\color{blue}{\left(B \cdot B + \left(-4\right) \cdot \left(A \cdot C\right)\right)} \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              4. metadata-eval17.4%

                \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{-4} \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              5. *-commutative17.4%

                \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{\left(A \cdot C\right) \cdot -4}\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              6. associate-*r*17.4%

                \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{A \cdot \left(C \cdot -4\right)}\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              7. fma-udef17.4%

                \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
            6. Applied egg-rr17.4%

              \[\leadsto \frac{-\color{blue}{1 \cdot \sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
            7. Step-by-step derivation
              1. *-lft-identity17.4%

                \[\leadsto \frac{-\color{blue}{\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              2. *-commutative17.4%

                \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(\left(F \cdot \left(A \cdot 2\right)\right) \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              3. associate-*r*17.4%

                \[\leadsto \frac{-\sqrt{2 \cdot \left(\color{blue}{\left(\left(F \cdot A\right) \cdot 2\right)} \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              4. *-commutative17.4%

                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{\left(A \cdot F\right)} \cdot 2\right) \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
              5. *-commutative17.4%

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

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

            if 2.70000000000000014e-31 < B

            1. Initial program 18.4%

              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. Step-by-step derivation
              1. Simplified18.4%

                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
              2. Taylor expanded in C around 0 25.6%

                \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
              3. Step-by-step derivation
                1. mul-1-neg25.6%

                  \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                2. *-commutative25.6%

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                3. +-commutative25.6%

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

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A} + {B}^{2}}\right)} \]
                5. unpow225.6%

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
                6. hypot-def51.9%

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
              4. Simplified51.9%

                \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
            3. Recombined 2 regimes into one program.
            4. Final simplification26.5%

              \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.7 \cdot 10^{-31}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(2 \cdot \left(A \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

            Alternative 3: 43.9% accurate, 2.8× speedup?

            \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 0.02:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(2 \cdot \left(A \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]
            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
             (if (<= B 0.02)
               (/
                (- (sqrt (* 2.0 (* (fma B B (* A (* C -4.0))) (* 2.0 (* A F))))))
                (- (* B B) (* 4.0 (* A C))))
               (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A B)))))))
            B = abs(B);
            assert(A < C);
            double code(double A, double B, double C, double F) {
            	double tmp;
            	if (B <= 0.02) {
            		tmp = -sqrt((2.0 * (fma(B, B, (A * (C * -4.0))) * (2.0 * (A * F))))) / ((B * B) - (4.0 * (A * C)));
            	} else {
            		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
            	}
            	return tmp;
            }
            
            B = abs(B)
            A, C = sort([A, C])
            function code(A, B, C, F)
            	tmp = 0.0
            	if (B <= 0.02)
            		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(fma(B, B, Float64(A * Float64(C * -4.0))) * Float64(2.0 * Float64(A * F)))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
            	else
            		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(F * Float64(A - 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_] := If[LessEqual[B, 0.02], N[((-N[Sqrt[N[(2.0 * N[(N[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(A * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]
            
            \begin{array}{l}
            B = |B|\\
            [A, C] = \mathsf{sort}([A, C])\\
            \\
            \begin{array}{l}
            \mathbf{if}\;B \leq 0.02:\\
            \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(2 \cdot \left(A \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\
            
            \mathbf{else}:\\
            \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\
            
            
            \end{array}
            \end{array}
            
            Derivation
            1. Split input into 2 regimes
            2. if B < 0.0200000000000000004

              1. Initial program 20.3%

                \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              2. Step-by-step derivation
                1. Simplified20.3%

                  \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                2. Taylor expanded in A around -inf 16.4%

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

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

                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                5. Step-by-step derivation
                  1. *-un-lft-identity16.4%

                    \[\leadsto \frac{-\color{blue}{1 \cdot \sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A \cdot 2\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  2. associate-*l*17.4%

                    \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \color{blue}{\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  3. cancel-sign-sub-inv17.4%

                    \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\color{blue}{\left(B \cdot B + \left(-4\right) \cdot \left(A \cdot C\right)\right)} \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  4. metadata-eval17.4%

                    \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{-4} \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  5. *-commutative17.4%

                    \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{\left(A \cdot C\right) \cdot -4}\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  6. associate-*r*17.4%

                    \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\left(B \cdot B + \color{blue}{A \cdot \left(C \cdot -4\right)}\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  7. fma-udef17.4%

                    \[\leadsto \frac{-1 \cdot \sqrt{2 \cdot \left(\color{blue}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                6. Applied egg-rr17.4%

                  \[\leadsto \frac{-\color{blue}{1 \cdot \sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                7. Step-by-step derivation
                  1. *-lft-identity17.4%

                    \[\leadsto \frac{-\color{blue}{\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A \cdot 2\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  2. *-commutative17.4%

                    \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(\left(F \cdot \left(A \cdot 2\right)\right) \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  3. associate-*r*17.4%

                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\color{blue}{\left(\left(F \cdot A\right) \cdot 2\right)} \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  4. *-commutative17.4%

                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{\left(A \cdot F\right)} \cdot 2\right) \cdot \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                  5. *-commutative17.4%

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

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

                if 0.0200000000000000004 < B

                1. Initial program 18.4%

                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                2. Step-by-step derivation
                  1. Simplified18.4%

                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                  2. Taylor expanded in C around 0 27.5%

                    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                  3. Step-by-step derivation
                    1. mul-1-neg27.5%

                      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                    2. *-commutative27.5%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                    3. +-commutative27.5%

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

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

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
                    6. hypot-def56.3%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
                  4. Simplified56.3%

                    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
                  5. Taylor expanded in A around 0 54.6%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A + -1 \cdot B\right)}} \]
                  6. Step-by-step derivation
                    1. mul-1-neg54.6%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A + \color{blue}{\left(-B\right)}\right)} \]
                    2. unsub-neg54.6%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}} \]
                  7. Simplified54.6%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}} \]
                3. Recombined 2 regimes into one program.
                4. Final simplification26.4%

                  \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 0.02:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(2 \cdot \left(A \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

                Alternative 4: 43.9% accurate, 3.0× speedup?

                \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 0.0125:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]
                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 0.0125)
                     (/ (- (sqrt (* 2.0 (* (* F t_0) (* 2.0 A))))) t_0)
                     (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A 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 <= 0.0125) {
                		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                	} else {
                		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
                	}
                	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) :: tmp
                    t_0 = (b * b) - (4.0d0 * (a * c))
                    if (b <= 0.0125d0) then
                        tmp = -sqrt((2.0d0 * ((f * t_0) * (2.0d0 * a)))) / t_0
                    else
                        tmp = (sqrt(2.0d0) / b) * -sqrt((f * (a - b)))
                    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 tmp;
                	if (B <= 0.0125) {
                		tmp = -Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                	} else {
                		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt((F * (A - 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 <= 0.0125:
                		tmp = -math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0
                	else:
                		tmp = (math.sqrt(2.0) / B) * -math.sqrt((F * (A - 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 <= 0.0125)
                		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A))))) / t_0);
                	else
                		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(F * Float64(A - 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 <= 0.0125)
                		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                	else
                		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - 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, 0.0125], 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[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $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 0.0125:\\
                \;\;\;\;\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}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\
                
                
                \end{array}
                \end{array}
                
                Derivation
                1. Split input into 2 regimes
                2. if B < 0.012500000000000001

                  1. Initial program 20.3%

                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                  2. Step-by-step derivation
                    1. Simplified20.3%

                      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                    2. Taylor expanded in A around -inf 16.4%

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

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

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

                    if 0.012500000000000001 < B

                    1. Initial program 18.4%

                      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                    2. Step-by-step derivation
                      1. Simplified18.4%

                        \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                      2. Taylor expanded in C around 0 27.5%

                        \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                      3. Step-by-step derivation
                        1. mul-1-neg27.5%

                          \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                        2. *-commutative27.5%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                        3. +-commutative27.5%

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

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

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
                        6. hypot-def56.3%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
                      4. Simplified56.3%

                        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
                      5. Taylor expanded in A around 0 54.6%

                        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A + -1 \cdot B\right)}} \]
                      6. Step-by-step derivation
                        1. mul-1-neg54.6%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A + \color{blue}{\left(-B\right)}\right)} \]
                        2. unsub-neg54.6%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}} \]
                      7. Simplified54.6%

                        \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}} \]
                    3. Recombined 2 regimes into one program.
                    4. Final simplification25.6%

                      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 0.0125:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

                    Alternative 5: 43.4% accurate, 3.0× speedup?

                    \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 0.013:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{-B \cdot F}\right)\\ \end{array} \end{array} \]
                    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 0.013)
                         (/ (- (sqrt (* 2.0 (* (* F t_0) (* 2.0 A))))) t_0)
                         (* (/ (sqrt 2.0) B) (- (sqrt (- (* B F))))))))
                    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 <= 0.013) {
                    		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                    	} else {
                    		tmp = (sqrt(2.0) / B) * -sqrt(-(B * F));
                    	}
                    	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) :: tmp
                        t_0 = (b * b) - (4.0d0 * (a * c))
                        if (b <= 0.013d0) then
                            tmp = -sqrt((2.0d0 * ((f * t_0) * (2.0d0 * a)))) / t_0
                        else
                            tmp = (sqrt(2.0d0) / b) * -sqrt(-(b * f))
                        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 tmp;
                    	if (B <= 0.013) {
                    		tmp = -Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                    	} else {
                    		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt(-(B * F));
                    	}
                    	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 <= 0.013:
                    		tmp = -math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0
                    	else:
                    		tmp = (math.sqrt(2.0) / B) * -math.sqrt(-(B * F))
                    	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 <= 0.013)
                    		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A))))) / t_0);
                    	else
                    		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(-Float64(B * F)))));
                    	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 <= 0.013)
                    		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                    	else
                    		tmp = (sqrt(2.0) / B) * -sqrt(-(B * F));
                    	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, 0.013], 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[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[(-N[(B * F), $MachinePrecision])], $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 0.013:\\
                    \;\;\;\;\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}}{B} \cdot \left(-\sqrt{-B \cdot F}\right)\\
                    
                    
                    \end{array}
                    \end{array}
                    
                    Derivation
                    1. Split input into 2 regimes
                    2. if B < 0.0129999999999999994

                      1. Initial program 20.3%

                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                      2. Step-by-step derivation
                        1. Simplified20.3%

                          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                        2. Taylor expanded in A around -inf 16.4%

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

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

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

                        if 0.0129999999999999994 < B

                        1. Initial program 18.4%

                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                        2. Step-by-step derivation
                          1. Simplified18.4%

                            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                          2. Taylor expanded in C around 0 27.5%

                            \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                          3. Step-by-step derivation
                            1. mul-1-neg27.5%

                              \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                            2. *-commutative27.5%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                            3. +-commutative27.5%

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

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

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)} \]
                            6. hypot-def56.3%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)} \]
                          4. Simplified56.3%

                            \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}} \]
                          5. Taylor expanded in A around 0 55.0%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-1 \cdot B\right)}} \]
                          6. Step-by-step derivation
                            1. mul-1-neg55.0%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
                          7. Simplified55.0%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \color{blue}{\left(-B\right)}} \]
                        3. Recombined 2 regimes into one program.
                        4. Final simplification25.7%

                          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 0.013:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{-B \cdot F}\right)\\ \end{array} \]

                        Alternative 6: 28.8% accurate, 4.8× speedup?

                        \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;B \leq 0.0118:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\ \end{array} \end{array} \]
                        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 0.0118)
                             (/ (- (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 <= 0.0118) {
                        		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 <= 0.0118d0) 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 <= 0.0118) {
                        		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 <= 0.0118:
                        		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 <= 0.0118)
                        		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 <= 0.0118)
                        		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, 0.0118], 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 0.0118:\\
                        \;\;\;\;\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 < 0.0117999999999999997

                          1. Initial program 20.3%

                            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                          2. Step-by-step derivation
                            1. Simplified20.3%

                              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                            2. Taylor expanded in A around -inf 16.4%

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

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

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

                            if 0.0117999999999999997 < B

                            1. Initial program 18.4%

                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                            2. Step-by-step derivation
                              1. Simplified18.4%

                                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                              2. Taylor expanded in B around inf 18.3%

                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \color{blue}{B}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                            3. Recombined 2 regimes into one program.
                            4. Final simplification16.9%

                              \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 0.0118:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                            Alternative 7: 29.7% accurate, 4.9× speedup?

                            \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 2.1 \cdot 10^{+50}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]
                            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 2.1e+50)
                                 (/ (- (sqrt (* 2.0 (* (* F t_0) (* 2.0 A))))) t_0)
                                 (* -2.0 (/ (sqrt (* A F)) 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 <= 2.1e+50) {
                            		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                            	} else {
                            		tmp = -2.0 * (sqrt((A * F)) / B);
                            	}
                            	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) :: tmp
                                t_0 = (b * b) - (4.0d0 * (a * c))
                                if (b <= 2.1d+50) then
                                    tmp = -sqrt((2.0d0 * ((f * t_0) * (2.0d0 * a)))) / t_0
                                else
                                    tmp = (-2.0d0) * (sqrt((a * f)) / b)
                                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 tmp;
                            	if (B <= 2.1e+50) {
                            		tmp = -Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                            	} else {
                            		tmp = -2.0 * (Math.sqrt((A * F)) / 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 <= 2.1e+50:
                            		tmp = -math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0
                            	else:
                            		tmp = -2.0 * (math.sqrt((A * F)) / 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 <= 2.1e+50)
                            		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A))))) / t_0);
                            	else
                            		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / 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 <= 2.1e+50)
                            		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                            	else
                            		tmp = -2.0 * (sqrt((A * F)) / 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, 2.1e+50], 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[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $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 2.1 \cdot 10^{+50}:\\
                            \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
                            
                            \mathbf{else}:\\
                            \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\
                            
                            
                            \end{array}
                            \end{array}
                            
                            Derivation
                            1. Split input into 2 regimes
                            2. if B < 2.1e50

                              1. Initial program 22.1%

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

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

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

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                4. Simplified15.7%

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

                                if 2.1e50 < B

                                1. Initial program 10.5%

                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                2. Step-by-step derivation
                                  1. Simplified10.5%

                                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                  2. Taylor expanded in A around -inf 2.7%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  3. Step-by-step derivation
                                    1. *-commutative2.7%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  4. Simplified2.7%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  5. Taylor expanded in B around inf 4.7%

                                    \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
                                  6. Step-by-step derivation
                                    1. associate-*r/4.8%

                                      \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]
                                    2. *-rgt-identity4.8%

                                      \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{A \cdot F}}}{B} \]
                                  7. Simplified4.8%

                                    \[\leadsto \color{blue}{-2 \cdot \frac{\sqrt{A \cdot F}}{B}} \]
                                3. Recombined 2 regimes into one program.
                                4. Final simplification13.6%

                                  \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.1 \cdot 10^{+50}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

                                Alternative 8: 24.1% accurate, 5.0× speedup?

                                \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 1.3 \cdot 10^{+51}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]
                                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
                                 (if (<= B 1.3e+51)
                                   (/
                                    (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* A (* C F)))))))
                                    (- (* B B) (* 4.0 (* A C))))
                                   (* -2.0 (/ (sqrt (* A F)) B))))
                                B = abs(B);
                                assert(A < C);
                                double code(double A, double B, double C, double F) {
                                	double tmp;
                                	if (B <= 1.3e+51) {
                                		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (A * C)));
                                	} else {
                                		tmp = -2.0 * (sqrt((A * F)) / B);
                                	}
                                	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) :: tmp
                                    if (b <= 1.3d+51) then
                                        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (a * (c * f)))))) / ((b * b) - (4.0d0 * (a * c)))
                                    else
                                        tmp = (-2.0d0) * (sqrt((a * f)) / b)
                                    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 tmp;
                                	if (B <= 1.3e+51) {
                                		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (A * C)));
                                	} else {
                                		tmp = -2.0 * (Math.sqrt((A * F)) / B);
                                	}
                                	return tmp;
                                }
                                
                                B = abs(B)
                                [A, C] = sort([A, C])
                                def code(A, B, C, F):
                                	tmp = 0
                                	if B <= 1.3e+51:
                                		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (A * C)))
                                	else:
                                		tmp = -2.0 * (math.sqrt((A * F)) / B)
                                	return tmp
                                
                                B = abs(B)
                                A, C = sort([A, C])
                                function code(A, B, C, F)
                                	tmp = 0.0
                                	if (B <= 1.3e+51)
                                		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(-4.0 * Float64(A * Float64(C * F))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
                                	else
                                		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
                                	end
                                	return tmp
                                end
                                
                                B = abs(B)
                                A, C = num2cell(sort([A, C])){:}
                                function tmp_2 = code(A, B, C, F)
                                	tmp = 0.0;
                                	if (B <= 1.3e+51)
                                		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (A * C)));
                                	else
                                		tmp = -2.0 * (sqrt((A * F)) / 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_] := If[LessEqual[B, 1.3e+51], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(-4.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]
                                
                                \begin{array}{l}
                                B = |B|\\
                                [A, C] = \mathsf{sort}([A, C])\\
                                \\
                                \begin{array}{l}
                                \mathbf{if}\;B \leq 1.3 \cdot 10^{+51}:\\
                                \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\
                                
                                \mathbf{else}:\\
                                \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\
                                
                                
                                \end{array}
                                \end{array}
                                
                                Derivation
                                1. Split input into 2 regimes
                                2. if B < 1.3000000000000001e51

                                  1. Initial program 22.1%

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

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

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

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    4. Simplified15.7%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    5. Taylor expanded in B around 0 13.0%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\color{blue}{\left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)} \cdot \left(A \cdot 2\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    6. Step-by-step derivation
                                      1. *-commutative13.0%

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(-4 \cdot \left(A \cdot \color{blue}{\left(F \cdot C\right)}\right)\right) \cdot \left(A \cdot 2\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    7. Simplified13.0%

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

                                    if 1.3000000000000001e51 < B

                                    1. Initial program 10.5%

                                      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                    2. Step-by-step derivation
                                      1. Simplified10.5%

                                        \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                      2. Taylor expanded in A around -inf 2.7%

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                      3. Step-by-step derivation
                                        1. *-commutative2.7%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                      4. Simplified2.7%

                                        \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                      5. Taylor expanded in B around inf 4.7%

                                        \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
                                      6. Step-by-step derivation
                                        1. associate-*r/4.8%

                                          \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]
                                        2. *-rgt-identity4.8%

                                          \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{A \cdot F}}}{B} \]
                                      7. Simplified4.8%

                                        \[\leadsto \color{blue}{-2 \cdot \frac{\sqrt{A \cdot F}}{B}} \]
                                    3. Recombined 2 regimes into one program.
                                    4. Final simplification11.4%

                                      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.3 \cdot 10^{+51}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

                                    Alternative 9: 27.6% accurate, 5.0× speedup?

                                    \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 1.85 \cdot 10^{+50}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(A \cdot \left(C \cdot -4\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]
                                    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
                                     (if (<= B 1.85e+50)
                                       (/
                                        (- (sqrt (* 2.0 (* (* 2.0 A) (* F (* A (* C -4.0)))))))
                                        (- (* B B) (* 4.0 (* A C))))
                                       (* -2.0 (/ (sqrt (* A F)) B))))
                                    B = abs(B);
                                    assert(A < C);
                                    double code(double A, double B, double C, double F) {
                                    	double tmp;
                                    	if (B <= 1.85e+50) {
                                    		tmp = -sqrt((2.0 * ((2.0 * A) * (F * (A * (C * -4.0)))))) / ((B * B) - (4.0 * (A * C)));
                                    	} else {
                                    		tmp = -2.0 * (sqrt((A * F)) / B);
                                    	}
                                    	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) :: tmp
                                        if (b <= 1.85d+50) then
                                            tmp = -sqrt((2.0d0 * ((2.0d0 * a) * (f * (a * (c * (-4.0d0))))))) / ((b * b) - (4.0d0 * (a * c)))
                                        else
                                            tmp = (-2.0d0) * (sqrt((a * f)) / b)
                                        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 tmp;
                                    	if (B <= 1.85e+50) {
                                    		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (F * (A * (C * -4.0)))))) / ((B * B) - (4.0 * (A * C)));
                                    	} else {
                                    		tmp = -2.0 * (Math.sqrt((A * F)) / B);
                                    	}
                                    	return tmp;
                                    }
                                    
                                    B = abs(B)
                                    [A, C] = sort([A, C])
                                    def code(A, B, C, F):
                                    	tmp = 0
                                    	if B <= 1.85e+50:
                                    		tmp = -math.sqrt((2.0 * ((2.0 * A) * (F * (A * (C * -4.0)))))) / ((B * B) - (4.0 * (A * C)))
                                    	else:
                                    		tmp = -2.0 * (math.sqrt((A * F)) / B)
                                    	return tmp
                                    
                                    B = abs(B)
                                    A, C = sort([A, C])
                                    function code(A, B, C, F)
                                    	tmp = 0.0
                                    	if (B <= 1.85e+50)
                                    		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(F * Float64(A * Float64(C * -4.0))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
                                    	else
                                    		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
                                    	end
                                    	return tmp
                                    end
                                    
                                    B = abs(B)
                                    A, C = num2cell(sort([A, C])){:}
                                    function tmp_2 = code(A, B, C, F)
                                    	tmp = 0.0;
                                    	if (B <= 1.85e+50)
                                    		tmp = -sqrt((2.0 * ((2.0 * A) * (F * (A * (C * -4.0)))))) / ((B * B) - (4.0 * (A * C)));
                                    	else
                                    		tmp = -2.0 * (sqrt((A * F)) / 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_] := If[LessEqual[B, 1.85e+50], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(F * N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]
                                    
                                    \begin{array}{l}
                                    B = |B|\\
                                    [A, C] = \mathsf{sort}([A, C])\\
                                    \\
                                    \begin{array}{l}
                                    \mathbf{if}\;B \leq 1.85 \cdot 10^{+50}:\\
                                    \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(A \cdot \left(C \cdot -4\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\
                                    
                                    \mathbf{else}:\\
                                    \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\
                                    
                                    
                                    \end{array}
                                    \end{array}
                                    
                                    Derivation
                                    1. Split input into 2 regimes
                                    2. if B < 1.85e50

                                      1. Initial program 22.1%

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

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

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

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        4. Simplified15.7%

                                          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        5. Taylor expanded in B around 0 15.3%

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

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{\left(\left(A \cdot C\right) \cdot -4\right)} \cdot F\right) \cdot \left(A \cdot 2\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          2. associate-*r*15.3%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\color{blue}{\left(A \cdot \left(C \cdot -4\right)\right)} \cdot F\right) \cdot \left(A \cdot 2\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                        7. Simplified15.3%

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

                                        if 1.85e50 < B

                                        1. Initial program 10.5%

                                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                        2. Step-by-step derivation
                                          1. Simplified10.5%

                                            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                          2. Taylor expanded in A around -inf 2.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          3. Step-by-step derivation
                                            1. *-commutative2.7%

                                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          4. Simplified2.7%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          5. Taylor expanded in B around inf 4.7%

                                            \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
                                          6. Step-by-step derivation
                                            1. associate-*r/4.8%

                                              \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]
                                            2. *-rgt-identity4.8%

                                              \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{A \cdot F}}}{B} \]
                                          7. Simplified4.8%

                                            \[\leadsto \color{blue}{-2 \cdot \frac{\sqrt{A \cdot F}}{B}} \]
                                        3. Recombined 2 regimes into one program.
                                        4. Final simplification13.2%

                                          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.85 \cdot 10^{+50}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(A \cdot \left(C \cdot -4\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

                                        Alternative 10: 9.2% accurate, 5.9× speedup?

                                        \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -2 \cdot \frac{\sqrt{A \cdot F}}{B} \end{array} \]
                                        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 (* A F)) B)))
                                        B = abs(B);
                                        assert(A < C);
                                        double code(double A, double B, double C, double F) {
                                        	return -2.0 * (sqrt((A * 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((a * 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((A * F)) / B);
                                        }
                                        
                                        B = abs(B)
                                        [A, C] = sort([A, C])
                                        def code(A, B, C, F):
                                        	return -2.0 * (math.sqrt((A * F)) / B)
                                        
                                        B = abs(B)
                                        A, C = sort([A, C])
                                        function code(A, B, C, F)
                                        	return Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B))
                                        end
                                        
                                        B = abs(B)
                                        A, C = num2cell(sort([A, C])){:}
                                        function tmp = code(A, B, C, F)
                                        	tmp = -2.0 * (sqrt((A * 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[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]
                                        
                                        \begin{array}{l}
                                        B = |B|\\
                                        [A, C] = \mathsf{sort}([A, C])\\
                                        \\
                                        -2 \cdot \frac{\sqrt{A \cdot F}}{B}
                                        \end{array}
                                        
                                        Derivation
                                        1. Initial program 19.8%

                                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                        2. Step-by-step derivation
                                          1. Simplified19.8%

                                            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                          2. Taylor expanded in A around -inf 13.1%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          3. Step-by-step derivation
                                            1. *-commutative13.1%

                                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          4. Simplified13.1%

                                            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                          5. Taylor expanded in B around inf 2.2%

                                            \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
                                          6. Step-by-step derivation
                                            1. associate-*r/2.2%

                                              \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]
                                            2. *-rgt-identity2.2%

                                              \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{A \cdot F}}}{B} \]
                                          7. Simplified2.2%

                                            \[\leadsto \color{blue}{-2 \cdot \frac{\sqrt{A \cdot F}}{B}} \]
                                          8. Final simplification2.2%

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

                                          Reproduce

                                          ?
                                          herbie shell --seed 2023200 
                                          (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))))