ABCF->ab-angle b

Percentage Accurate: 19.2% → 48.0%
Time: 23.8s
Alternatives: 10
Speedup: 5.1×

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: 19.2% 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: 48.0% accurate, 0.3× speedup?

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\\ t_2 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ t_3 := \frac{-\sqrt{\left(2 \cdot \left(t_2 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{t_2}\\ \mathbf{if}\;t_3 \leq -5 \cdot 10^{-172}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{t_1}\\ \mathbf{elif}\;t_3 \leq 2 \cdot 10^{-89}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{t_0}\\ \mathbf{elif}\;t_3 \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(A + \left(A - -0.5 \cdot \left(B \cdot \frac{B}{A}\right)\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, B \cdot \left(2 \cdot B\right)\right)} \cdot \left(-\sqrt{F}\right)}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)}\right)\\ \end{array} \end{array} \]
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 (* t_2 F))
             (- (+ A C) (sqrt (+ (pow B 2.0) (pow (- A C) 2.0)))))))
          t_2)))
   (if (<= t_3 -5e-172)
     (/
      (*
       (sqrt (* 2.0 (* F (+ A (- C (hypot (- A C) B))))))
       (- (sqrt (fma B B (* C (* A -4.0))))))
      t_1)
     (if (<= t_3 2e-89)
       (/
        (-
         (sqrt
          (*
           2.0
           (*
            (* F t_0)
            (+ A (+ A (* -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C))))))))
        t_0)
       (if (<= t_3 INFINITY)
         (/
          (*
           (sqrt
            (*
             (+ A (- A (* -0.5 (* B (/ B A)))))
             (fma C (* A -8.0) (* B (* 2.0 B)))))
           (- (sqrt F)))
          t_1)
         (* (/ (sqrt 2.0) B) (- (sqrt (* F (- C (hypot C 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 * (t_2 * F)) * ((A + C) - sqrt((pow(B, 2.0) + pow((A - C), 2.0)))))) / t_2;
	double tmp;
	if (t_3 <= -5e-172) {
		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 <= 2e-89) {
		tmp = -sqrt((2.0 * ((F * t_0) * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	} else if (t_3 <= ((double) INFINITY)) {
		tmp = (sqrt(((A + (A - (-0.5 * (B * (B / A))))) * fma(C, (A * -8.0), (B * (2.0 * B))))) * -sqrt(F)) / t_1;
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((F * (C - hypot(C, 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(t_2 * F)) * Float64(Float64(A + C) - sqrt(Float64((B ^ 2.0) + (Float64(A - C) ^ 2.0))))))) / t_2)
	tmp = 0.0
	if (t_3 <= -5e-172)
		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 <= 2e-89)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + Float64(A + Float64(-0.5 * Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C)))))))) / t_0);
	elseif (t_3 <= Inf)
		tmp = Float64(Float64(sqrt(Float64(Float64(A + Float64(A - Float64(-0.5 * Float64(B * Float64(B / A))))) * 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(C - hypot(C, 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[(t$95$2 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[B, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$2), $MachinePrecision]}, If[LessEqual[t$95$3, -5e-172], 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, 2e-89], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[t$95$3, Infinity], N[(N[(N[Sqrt[N[(N[(A + N[(A - N[(-0.5 * N[(B * N[(B / A), $MachinePrecision]), $MachinePrecision]), $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[(C - N[Sqrt[C ^ 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(t_2 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{t_2}\\
\mathbf{if}\;t_3 \leq -5 \cdot 10^{-172}:\\
\;\;\;\;\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 2 \cdot 10^{-89}:\\
\;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{t_0}\\

\mathbf{elif}\;t_3 \leq \infty:\\
\;\;\;\;\frac{\sqrt{\left(A + \left(A - -0.5 \cdot \left(B \cdot \frac{B}{A}\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(C - \mathsf{hypot}\left(C, 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))) < -4.9999999999999999e-172

    1. Initial program 42.7%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Step-by-step derivation
      1. Simplified54.0%

        \[\leadsto \color{blue}{\frac{-\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)}} \]
      2. Step-by-step derivation
        1. sqrt-prod63.8%

          \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        2. associate-*r*63.8%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, \color{blue}{\left(A \cdot -4\right) \cdot C}\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        3. *-commutative63.8%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, \color{blue}{C \cdot \left(A \cdot -4\right)}\right)} \cdot \sqrt{\left(2 \cdot F\right) \cdot \left(C - \left(\mathsf{hypot}\left(B, A - C\right) - A\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        4. associate-*l*63.8%

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

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

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

        \[\leadsto \frac{-\color{blue}{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
      4. Step-by-step derivation
        1. hypot-def47.6%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \color{blue}{\sqrt{B \cdot B + \left(A - C\right) \cdot \left(A - C\right)}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        2. unpow247.6%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{B \cdot B + \color{blue}{{\left(A - C\right)}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        3. unpow247.6%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{{B}^{2}} + {\left(A - C\right)}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        4. +-commutative47.6%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{{\left(A - C\right)}^{2} + {B}^{2}}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        5. unpow247.6%

          \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \sqrt{\color{blue}{\left(A - C\right) \cdot \left(A - C\right)} + {B}^{2}}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(-4 \cdot C\right)\right)} \]
        6. unpow247.6%

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

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

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

      if -4.9999999999999999e-172 < (/.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))) < 2.00000000000000008e-89

      1. Initial program 8.6%

        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. Step-by-step derivation
        1. Simplified8.6%

          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
        2. Taylor expanded in C around inf 18.1%

          \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        3. Step-by-step derivation
          1. associate--l+18.1%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} - -1 \cdot A\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          2. associate--l+18.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          3. unpow218.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          4. unpow218.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          5. unpow218.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          6. mul-1-neg18.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          7. mul-1-neg18.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          8. sqr-neg18.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
          9. mul-1-neg18.4%

            \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
        4. Simplified18.4%

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

        if 2.00000000000000008e-89 < (/.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 35.6%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Simplified40.0%

          \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
        3. Taylor expanded in A around -inf 27.1%

          \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{A} + -1 \cdot A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        4. Step-by-step derivation
          1. fma-def27.1%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\mathsf{fma}\left(-0.5, \frac{{B}^{2}}{A}, -1 \cdot A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          2. unpow227.1%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(-0.5, \frac{\color{blue}{B \cdot B}}{A}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          3. mul-1-neg27.1%

            \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(-0.5, \frac{B \cdot B}{A}, \color{blue}{-A}\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        5. Simplified27.1%

          \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\mathsf{fma}\left(-0.5, \frac{B \cdot B}{A}, -A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        6. Step-by-step derivation
          1. sqrt-prod41.3%

            \[\leadsto \frac{-\color{blue}{\sqrt{F} \cdot \sqrt{\left(A - \mathsf{fma}\left(-0.5, \frac{B \cdot B}{A}, -A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          2. associate-/l*41.3%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\left(A - \mathsf{fma}\left(-0.5, \color{blue}{\frac{B}{\frac{A}{B}}}, -A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        7. Applied egg-rr41.3%

          \[\leadsto \frac{-\color{blue}{\sqrt{F} \cdot \sqrt{\left(A - \mathsf{fma}\left(-0.5, \frac{B}{\frac{A}{B}}, -A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        8. Step-by-step derivation
          1. fma-udef41.3%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\left(A - \color{blue}{\left(-0.5 \cdot \frac{B}{\frac{A}{B}} + \left(-A\right)\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          2. unsub-neg41.3%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\left(A - \color{blue}{\left(-0.5 \cdot \frac{B}{\frac{A}{B}} - A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          3. associate-/r/41.3%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\left(A - \left(-0.5 \cdot \color{blue}{\left(\frac{B}{A} \cdot B\right)} - A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
          4. associate-*r*41.3%

            \[\leadsto \frac{-\sqrt{F} \cdot \sqrt{\left(A - \left(-0.5 \cdot \left(\frac{B}{A} \cdot B\right) - A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, \color{blue}{\left(2 \cdot B\right) \cdot B}\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
        9. Simplified41.3%

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

        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 A around 0 1.9%

            \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F}\right)} \]
          3. Step-by-step derivation
            1. mul-1-neg1.9%

              \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(C - \sqrt{{B}^{2} + {C}^{2}}\right) \cdot F}} \]
            2. *-commutative1.9%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(C - \sqrt{{B}^{2} + {C}^{2}}\right)}} \]
            3. +-commutative1.9%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{{C}^{2} + {B}^{2}}}\right)} \]
            4. unpow21.9%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{\color{blue}{C \cdot C} + {B}^{2}}\right)} \]
            5. unpow21.9%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \sqrt{C \cdot C + \color{blue}{B \cdot B}}\right)} \]
            6. hypot-def18.5%

              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \color{blue}{\mathsf{hypot}\left(C, B\right)}\right)} \]
          4. Simplified18.5%

            \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)}} \]
        3. Recombined 4 regimes into one program.
        4. Final simplification33.9%

          \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq -5 \cdot 10^{-172}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq 2 \cdot 10^{-89}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(A + \left(A - -0.5 \cdot \left(B \cdot \frac{B}{A}\right)\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, B \cdot \left(2 \cdot B\right)\right)} \cdot \left(-\sqrt{F}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)}\right)\\ \end{array} \]

        Alternative 2: 47.6% accurate, 3.0× speedup?

        \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 1180:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\\ \end{array} \end{array} \]
        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 1180.0)
             (/ (- (sqrt (* 2.0 (* (* F t_0) (* 2.0 A))))) t_0)
             (/ (- (sqrt (* (* 2.0 F) (- A (hypot B 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 <= 1180.0) {
        		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
        	} else {
        		tmp = -sqrt(((2.0 * F) * (A - hypot(B, A)))) / B;
        	}
        	return tmp;
        }
        
        B = Math.abs(B);
        assert A < C;
        public static double code(double A, double B, double C, double F) {
        	double t_0 = (B * B) - (4.0 * (A * C));
        	double tmp;
        	if (B <= 1180.0) {
        		tmp = -Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
        	} else {
        		tmp = -Math.sqrt(((2.0 * F) * (A - Math.hypot(B, 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 <= 1180.0:
        		tmp = -math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0
        	else:
        		tmp = -math.sqrt(((2.0 * F) * (A - math.hypot(B, 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 <= 1180.0)
        		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A))))) / t_0);
        	else
        		tmp = Float64(Float64(-sqrt(Float64(Float64(2.0 * F) * Float64(A - hypot(B, 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 <= 1180.0)
        		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
        	else
        		tmp = -sqrt(((2.0 * F) * (A - hypot(B, 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, 1180.0], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(N[(2.0 * F), $MachinePrecision] * N[(A - N[Sqrt[B ^ 2 + A ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / B), $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 1180:\\
        \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
        
        \mathbf{else}:\\
        \;\;\;\;\frac{-\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if B < 1180

          1. Initial program 19.1%

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

              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
            2. Taylor expanded in A around -inf 12.2%

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

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

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

            if 1180 < B

            1. Initial program 11.6%

              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. Step-by-step derivation
              1. Simplified11.6%

                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
              2. Taylor expanded in C around 0 17.0%

                \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
              3. Step-by-step derivation
                1. mul-1-neg17.0%

                  \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                2. *-commutative17.0%

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                3. unpow217.0%

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
                4. unpow217.0%

                  \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
              4. Simplified17.0%

                \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + A \cdot A}\right)}} \]
              5. Step-by-step derivation
                1. associate-*l/17.0%

                  \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + A \cdot A}\right)}}{B}} \]
                2. +-commutative17.0%

                  \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A + B \cdot B}}\right)}}{B} \]
                3. hypot-udef48.3%

                  \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)}}{B} \]
                4. hypot-udef17.0%

                  \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\sqrt{A \cdot A + B \cdot B}}\right)}}{B} \]
                5. +-commutative17.0%

                  \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B + A \cdot A}}\right)}}{B} \]
                6. hypot-def48.3%

                  \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)}}{B} \]
              6. Applied egg-rr48.3%

                \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}} \]
              7. Step-by-step derivation
                1. pow148.3%

                  \[\leadsto -\frac{\color{blue}{{\left(\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}\right)}^{1}}}{B} \]
                2. sqrt-unprod48.5%

                  \[\leadsto -\frac{{\color{blue}{\left(\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}\right)}}^{1}}{B} \]
              8. Applied egg-rr48.5%

                \[\leadsto -\frac{\color{blue}{{\left(\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}\right)}^{1}}}{B} \]
              9. Step-by-step derivation
                1. unpow148.5%

                  \[\leadsto -\frac{\color{blue}{\sqrt{2 \cdot \left(F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)\right)}}}{B} \]
                2. associate-*r*48.5%

                  \[\leadsto -\frac{\sqrt{\color{blue}{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}}{B} \]
              10. Simplified48.5%

                \[\leadsto -\frac{\color{blue}{\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}}{B} \]
            3. Recombined 2 regimes into one program.
            4. Final simplification21.1%

              \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1180:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{\left(2 \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}\\ \end{array} \]

            Alternative 3: 44.3% accurate, 3.0× speedup?

            \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 4600:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]
            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 4600.0)
                 (/ (- (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 <= 4600.0) {
            		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 <= 4600.0d0) 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 <= 4600.0) {
            		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 <= 4600.0:
            		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 <= 4600.0)
            		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 <= 4600.0)
            		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, 4600.0], 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 4600:\\
            \;\;\;\;\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 < 4600

              1. Initial program 19.1%

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

                  \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                2. Taylor expanded in A around -inf 12.2%

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

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

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

                if 4600 < B

                1. Initial program 11.6%

                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                2. Step-by-step derivation
                  1. Simplified11.6%

                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                  2. Taylor expanded in C around 0 17.0%

                    \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                  3. Step-by-step derivation
                    1. mul-1-neg17.0%

                      \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                    2. *-commutative17.0%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                    3. unpow217.0%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
                    4. unpow217.0%

                      \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
                  4. Simplified17.0%

                    \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + A \cdot A}\right)}} \]
                  5. Taylor expanded in B around inf 44.0%

                    \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{B}\right)} \]
                3. Recombined 2 regimes into one program.
                4. Final simplification20.0%

                  \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 4600:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

                Alternative 4: 44.2% accurate, 3.0× speedup?

                \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 780:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)}{B}\\ \end{array} \end{array} \]
                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 780.0)
                     (/ (- (sqrt (* 2.0 (* (* F t_0) (* 2.0 A))))) t_0)
                     (/ (* (sqrt 2.0) (- (sqrt (* F (- A B))))) 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 <= 780.0) {
                		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                	} else {
                		tmp = (sqrt(2.0) * -sqrt((F * (A - B)))) / 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 <= 780.0d0) then
                        tmp = -sqrt((2.0d0 * ((f * t_0) * (2.0d0 * a)))) / t_0
                    else
                        tmp = (sqrt(2.0d0) * -sqrt((f * (a - b)))) / 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 <= 780.0) {
                		tmp = -Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                	} else {
                		tmp = (Math.sqrt(2.0) * -Math.sqrt((F * (A - B)))) / 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 <= 780.0:
                		tmp = -math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0
                	else:
                		tmp = (math.sqrt(2.0) * -math.sqrt((F * (A - B)))) / 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 <= 780.0)
                		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) * Float64(-sqrt(Float64(F * Float64(A - B))))) / 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 <= 780.0)
                		tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                	else
                		tmp = (sqrt(2.0) * -sqrt((F * (A - B)))) / 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, 780.0], 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] * (-N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / B), $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 780:\\
                \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\
                
                \mathbf{else}:\\
                \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)}{B}\\
                
                
                \end{array}
                \end{array}
                
                Derivation
                1. Split input into 2 regimes
                2. if B < 780

                  1. Initial program 19.1%

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

                      \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                    2. Taylor expanded in A around -inf 12.2%

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

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

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

                    if 780 < B

                    1. Initial program 11.6%

                      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                    2. Step-by-step derivation
                      1. Simplified11.6%

                        \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                      2. Taylor expanded in C around 0 17.0%

                        \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                      3. Step-by-step derivation
                        1. mul-1-neg17.0%

                          \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                        2. *-commutative17.0%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                        3. unpow217.0%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
                        4. unpow217.0%

                          \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
                      4. Simplified17.0%

                        \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + A \cdot A}\right)}} \]
                      5. Step-by-step derivation
                        1. associate-*l/17.0%

                          \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + A \cdot A}\right)}}{B}} \]
                        2. +-commutative17.0%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{A \cdot A + B \cdot B}}\right)}}{B} \]
                        3. hypot-udef48.3%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)}}{B} \]
                        4. hypot-udef17.0%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\sqrt{A \cdot A + B \cdot B}}\right)}}{B} \]
                        5. +-commutative17.0%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B + A \cdot A}}\right)}}{B} \]
                        6. hypot-def48.3%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \color{blue}{\mathsf{hypot}\left(B, A\right)}\right)}}{B} \]
                      6. Applied egg-rr48.3%

                        \[\leadsto -\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A - \mathsf{hypot}\left(B, A\right)\right)}}{B}} \]
                      7. Taylor expanded in A around 0 44.0%

                        \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(A + -1 \cdot B\right)}}}{B} \]
                      8. Step-by-step derivation
                        1. mul-1-neg44.0%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(A + \color{blue}{\left(-B\right)}\right)}}{B} \]
                        2. sub-neg44.0%

                          \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}}}{B} \]
                      9. Simplified44.0%

                        \[\leadsto -\frac{\sqrt{2} \cdot \sqrt{F \cdot \color{blue}{\left(A - B\right)}}}{B} \]
                    3. Recombined 2 regimes into one program.
                    4. Final simplification20.0%

                      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 780:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)}{B}\\ \end{array} \]

                    Alternative 5: 43.7% accurate, 3.0× speedup?

                    \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 850:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{B \cdot \left(-F\right)}\right)\\ \end{array} \end{array} \]
                    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 850.0)
                         (/ (- (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 <= 850.0) {
                    		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 <= 850.0d0) 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 <= 850.0) {
                    		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 <= 850.0:
                    		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 <= 850.0)
                    		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(B * Float64(-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 <= 850.0)
                    		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, 850.0], 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 850:\\
                    \;\;\;\;\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 \left(-F\right)}\right)\\
                    
                    
                    \end{array}
                    \end{array}
                    
                    Derivation
                    1. Split input into 2 regimes
                    2. if B < 850

                      1. Initial program 19.1%

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

                          \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                        2. Taylor expanded in A around -inf 12.2%

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

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

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

                        if 850 < B

                        1. Initial program 11.6%

                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                        2. Step-by-step derivation
                          1. Simplified11.6%

                            \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                          2. Taylor expanded in C around 0 17.0%

                            \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}\right)} \]
                          3. Step-by-step derivation
                            1. mul-1-neg17.0%

                              \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{\left(A - \sqrt{{B}^{2} + {A}^{2}}\right) \cdot F}} \]
                            2. *-commutative17.0%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(A - \sqrt{{B}^{2} + {A}^{2}}\right)}} \]
                            3. unpow217.0%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{\color{blue}{B \cdot B} + {A}^{2}}\right)} \]
                            4. unpow217.0%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + \color{blue}{A \cdot A}}\right)} \]
                          4. Simplified17.0%

                            \[\leadsto \color{blue}{-\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \sqrt{B \cdot B + A \cdot A}\right)}} \]
                          5. Taylor expanded in B around inf 44.0%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(A - \color{blue}{B}\right)} \]
                          6. Taylor expanded in A around 0 43.8%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{-1 \cdot \left(F \cdot B\right)}} \]
                          7. Step-by-step derivation
                            1. mul-1-neg43.8%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{-F \cdot B}} \]
                            2. distribute-rgt-neg-in43.8%

                              \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(-B\right)}} \]
                          8. Simplified43.8%

                            \[\leadsto -\frac{\sqrt{2}}{B} \cdot \sqrt{\color{blue}{F \cdot \left(-B\right)}} \]
                        3. Recombined 2 regimes into one program.
                        4. Final simplification20.0%

                          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 850:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{B \cdot \left(-F\right)}\right)\\ \end{array} \]

                        Alternative 6: 29.9% accurate, 4.3× speedup?

                        \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;C \leq 3 \cdot 10^{-13}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]
                        NOTE: B should be positive before calling this function
                        NOTE: A and C should be sorted in increasing order before calling this function.
                        (FPCore (A B C F)
                         :precision binary64
                         (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* F t_0)))
                           (if (<= C 3e-13)
                             (/ (- (sqrt (* 2.0 (* t_1 (* 2.0 A))))) t_0)
                             (/
                              (-
                               (sqrt
                                (*
                                 2.0
                                 (* t_1 (+ A (+ A (* -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C))))))))
                              t_0))))
                        B = abs(B);
                        assert(A < C);
                        double code(double A, double B, double C, double F) {
                        	double t_0 = (B * B) - (4.0 * (A * C));
                        	double t_1 = F * t_0;
                        	double tmp;
                        	if (C <= 3e-13) {
                        		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
                        	} else {
                        		tmp = -sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / 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 (c <= 3d-13) then
                                tmp = -sqrt((2.0d0 * (t_1 * (2.0d0 * a)))) / t_0
                            else
                                tmp = -sqrt((2.0d0 * (t_1 * (a + (a + ((-0.5d0) * (((b * b) + ((a * a) - (a * a))) / c))))))) / 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 (C <= 3e-13) {
                        		tmp = -Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
                        	} else {
                        		tmp = -Math.sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
                        	}
                        	return tmp;
                        }
                        
                        B = abs(B)
                        [A, C] = sort([A, C])
                        def code(A, B, C, F):
                        	t_0 = (B * B) - (4.0 * (A * C))
                        	t_1 = F * t_0
                        	tmp = 0
                        	if C <= 3e-13:
                        		tmp = -math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0
                        	else:
                        		tmp = -math.sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0
                        	return tmp
                        
                        B = abs(B)
                        A, C = sort([A, C])
                        function code(A, B, C, F)
                        	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                        	t_1 = Float64(F * t_0)
                        	tmp = 0.0
                        	if (C <= 3e-13)
                        		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(A + Float64(A + Float64(-0.5 * Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C)))))))) / t_0);
                        	end
                        	return tmp
                        end
                        
                        B = abs(B)
                        A, C = num2cell(sort([A, C])){:}
                        function tmp_2 = code(A, B, C, F)
                        	t_0 = (B * B) - (4.0 * (A * C));
                        	t_1 = F * t_0;
                        	tmp = 0.0;
                        	if (C <= 3e-13)
                        		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
                        	else
                        		tmp = -sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
                        	end
                        	tmp_2 = tmp;
                        end
                        
                        NOTE: B should be positive before calling this function
                        NOTE: A and C should be sorted in increasing order before calling this function.
                        code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, If[LessEqual[C, 3e-13], 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[(A + N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]
                        
                        \begin{array}{l}
                        B = |B|\\
                        [A, C] = \mathsf{sort}([A, C])\\
                        \\
                        \begin{array}{l}
                        t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                        t_1 := F \cdot t_0\\
                        \mathbf{if}\;C \leq 3 \cdot 10^{-13}:\\
                        \;\;\;\;\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(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{t_0}\\
                        
                        
                        \end{array}
                        \end{array}
                        
                        Derivation
                        1. Split input into 2 regimes
                        2. if C < 2.99999999999999984e-13

                          1. Initial program 22.2%

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

                              \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                            2. Taylor expanded in A around -inf 11.6%

                              \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                            3. Step-by-step derivation
                              1. *-commutative11.6%

                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                            4. Simplified11.6%

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

                            if 2.99999999999999984e-13 < C

                            1. Initial program 2.7%

                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                            2. Step-by-step derivation
                              1. Simplified2.7%

                                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                              2. Taylor expanded in C around inf 21.0%

                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(\left(A + -0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C}\right) - -1 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                              3. Step-by-step derivation
                                1. associate--l+21.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{\left({B}^{2} + {A}^{2}\right) - {\left(-1 \cdot A\right)}^{2}}{C} - -1 \cdot A\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                2. associate--l+21.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{\color{blue}{{B}^{2} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                3. unpow221.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{\color{blue}{B \cdot B} + \left({A}^{2} - {\left(-1 \cdot A\right)}^{2}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                4. unpow221.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(\color{blue}{A \cdot A} - {\left(-1 \cdot A\right)}^{2}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                5. unpow221.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-1 \cdot A\right) \cdot \left(-1 \cdot A\right)}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                6. mul-1-neg21.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \color{blue}{\left(-A\right)} \cdot \left(-1 \cdot A\right)\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                7. mul-1-neg21.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \left(-A\right) \cdot \color{blue}{\left(-A\right)}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                8. sqr-neg21.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - \color{blue}{A \cdot A}\right)}{C} - -1 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                9. mul-1-neg21.0%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C} - \color{blue}{\left(-A\right)}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                              4. Simplified21.0%

                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A + \left(-0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C} - \left(-A\right)\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                            3. Recombined 2 regimes into one program.
                            4. Final simplification14.0%

                              \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 3 \cdot 10^{-13}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                            Alternative 7: 27.0% accurate, 5.0× speedup?

                            \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0} \end{array} \end{array} \]
                            NOTE: B should be positive before calling this function
                            NOTE: A and C should be sorted in increasing order before calling this function.
                            (FPCore (A B C F)
                             :precision binary64
                             (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
                               (/ (- (sqrt (* 2.0 (* (* F t_0) (* 2.0 A))))) t_0)))
                            B = abs(B);
                            assert(A < C);
                            double code(double A, double B, double C, double F) {
                            	double t_0 = (B * B) - (4.0 * (A * C));
                            	return -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                            }
                            
                            NOTE: B should be positive before calling this function
                            NOTE: A and C should be sorted in increasing order before calling this function.
                            real(8) function code(a, b, c, f)
                                real(8), intent (in) :: a
                                real(8), intent (in) :: b
                                real(8), intent (in) :: c
                                real(8), intent (in) :: f
                                real(8) :: t_0
                                t_0 = (b * b) - (4.0d0 * (a * c))
                                code = -sqrt((2.0d0 * ((f * t_0) * (2.0d0 * a)))) / t_0
                            end function
                            
                            B = Math.abs(B);
                            assert A < C;
                            public static double code(double A, double B, double C, double F) {
                            	double t_0 = (B * B) - (4.0 * (A * C));
                            	return -Math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                            }
                            
                            B = abs(B)
                            [A, C] = sort([A, C])
                            def code(A, B, C, F):
                            	t_0 = (B * B) - (4.0 * (A * C))
                            	return -math.sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0
                            
                            B = abs(B)
                            A, C = sort([A, C])
                            function code(A, B, C, F)
                            	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
                            	return Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(2.0 * A))))) / t_0)
                            end
                            
                            B = abs(B)
                            A, C = num2cell(sort([A, C])){:}
                            function tmp = code(A, B, C, F)
                            	t_0 = (B * B) - (4.0 * (A * C));
                            	tmp = -sqrt((2.0 * ((F * t_0) * (2.0 * A)))) / t_0;
                            end
                            
                            NOTE: B should be positive before calling this function
                            NOTE: A and C should be sorted in increasing order before calling this function.
                            code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]
                            
                            \begin{array}{l}
                            B = |B|\\
                            [A, C] = \mathsf{sort}([A, C])\\
                            \\
                            \begin{array}{l}
                            t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\
                            \frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}
                            \end{array}
                            \end{array}
                            
                            Derivation
                            1. Initial program 17.3%

                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                            2. Step-by-step derivation
                              1. Simplified17.3%

                                \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                              2. Taylor expanded in A around -inf 10.6%

                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(2 \cdot A\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                              3. Step-by-step derivation
                                1. *-commutative10.6%

                                  \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                              4. Simplified10.6%

                                \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A \cdot 2\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                              5. Final simplification10.6%

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

                              Alternative 8: 19.8% accurate, 5.1× speedup?

                              \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;A \leq -3.8 \cdot 10^{+155}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \end{array} \]
                              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 (<= A -3.8e+155)
                                 (* -2.0 (/ (sqrt (* A F)) B))
                                 (/
                                  (- (sqrt (* 2.0 (* -8.0 (* (* A A) (* C F))))))
                                  (- (* B B) (* 4.0 (* A C))))))
                              B = abs(B);
                              assert(A < C);
                              double code(double A, double B, double C, double F) {
                              	double tmp;
                              	if (A <= -3.8e+155) {
                              		tmp = -2.0 * (sqrt((A * F)) / B);
                              	} else {
                              		tmp = -sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)));
                              	}
                              	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 (a <= (-3.8d+155)) then
                                      tmp = (-2.0d0) * (sqrt((a * f)) / b)
                                  else
                                      tmp = -sqrt((2.0d0 * ((-8.0d0) * ((a * a) * (c * f))))) / ((b * b) - (4.0d0 * (a * c)))
                                  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 (A <= -3.8e+155) {
                              		tmp = -2.0 * (Math.sqrt((A * F)) / B);
                              	} else {
                              		tmp = -Math.sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)));
                              	}
                              	return tmp;
                              }
                              
                              B = abs(B)
                              [A, C] = sort([A, C])
                              def code(A, B, C, F):
                              	tmp = 0
                              	if A <= -3.8e+155:
                              		tmp = -2.0 * (math.sqrt((A * F)) / B)
                              	else:
                              		tmp = -math.sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)))
                              	return tmp
                              
                              B = abs(B)
                              A, C = sort([A, C])
                              function code(A, B, C, F)
                              	tmp = 0.0
                              	if (A <= -3.8e+155)
                              		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
                              	else
                              		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(-8.0 * Float64(Float64(A * A) * Float64(C * F)))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
                              	end
                              	return tmp
                              end
                              
                              B = abs(B)
                              A, C = num2cell(sort([A, C])){:}
                              function tmp_2 = code(A, B, C, F)
                              	tmp = 0.0;
                              	if (A <= -3.8e+155)
                              		tmp = -2.0 * (sqrt((A * F)) / B);
                              	else
                              		tmp = -sqrt((2.0 * (-8.0 * ((A * A) * (C * F))))) / ((B * B) - (4.0 * (A * C)));
                              	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[A, -3.8e+155], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(-8.0 * N[(N[(A * A), $MachinePrecision] * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]
                              
                              \begin{array}{l}
                              B = |B|\\
                              [A, C] = \mathsf{sort}([A, C])\\
                              \\
                              \begin{array}{l}
                              \mathbf{if}\;A \leq -3.8 \cdot 10^{+155}:\\
                              \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\
                              
                              \mathbf{else}:\\
                              \;\;\;\;\frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\
                              
                              
                              \end{array}
                              \end{array}
                              
                              Derivation
                              1. Split input into 2 regimes
                              2. if A < -3.8000000000000001e155

                                1. Initial program 1.8%

                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                2. Simplified9.8%

                                  \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
                                3. Taylor expanded in A around -inf 1.3%

                                  \[\leadsto \frac{-\sqrt{\color{blue}{-16 \cdot \left({A}^{2} \cdot \left(C \cdot F\right)\right) + 4 \cdot \left(A \cdot \left(F \cdot {B}^{2}\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                4. Step-by-step derivation
                                  1. fma-def1.3%

                                    \[\leadsto \frac{-\sqrt{\color{blue}{\mathsf{fma}\left(-16, {A}^{2} \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(F \cdot {B}^{2}\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  2. unpow21.3%

                                    \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(-16, \color{blue}{\left(A \cdot A\right)} \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(F \cdot {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  3. *-commutative1.3%

                                    \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(-16, \left(A \cdot A\right) \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \color{blue}{\left({B}^{2} \cdot F\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  4. unpow21.3%

                                    \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(-16, \left(A \cdot A\right) \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(\color{blue}{\left(B \cdot B\right)} \cdot F\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                5. Simplified1.3%

                                  \[\leadsto \frac{-\sqrt{\color{blue}{\mathsf{fma}\left(-16, \left(A \cdot A\right) \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(\left(B \cdot B\right) \cdot F\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                6. Taylor expanded in C around 0 9.8%

                                  \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
                                7. Step-by-step derivation
                                  1. associate-*r/9.8%

                                    \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]
                                  2. *-rgt-identity9.8%

                                    \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{A \cdot F}}}{B} \]
                                  3. *-commutative9.8%

                                    \[\leadsto -2 \cdot \frac{\sqrt{\color{blue}{F \cdot A}}}{B} \]
                                8. Simplified9.8%

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

                                if -3.8000000000000001e155 < A

                                1. Initial program 18.9%

                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                2. Step-by-step derivation
                                  1. Simplified18.9%

                                    \[\leadsto \color{blue}{\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - \sqrt{B \cdot B + {\left(A - C\right)}^{2}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}} \]
                                  2. Taylor expanded in C around 0 15.6%

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

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

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

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \sqrt{A \cdot A + \color{blue}{B \cdot B}}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    4. hypot-def15.8%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \color{blue}{\mathsf{hypot}\left(A, B\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  4. Simplified15.8%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \color{blue}{\left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  5. Taylor expanded in A around -inf 7.5%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(-8 \cdot \left({A}^{2} \cdot \left(C \cdot F\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  6. Step-by-step derivation
                                    1. *-commutative7.5%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \color{blue}{\left(\left(C \cdot F\right) \cdot {A}^{2}\right)}\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    2. *-commutative7.5%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\color{blue}{\left(F \cdot C\right)} \cdot {A}^{2}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                    3. unpow27.5%

                                      \[\leadsto \frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(F \cdot C\right) \cdot \color{blue}{\left(A \cdot A\right)}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                  7. Simplified7.5%

                                    \[\leadsto \frac{-\sqrt{2 \cdot \color{blue}{\left(-8 \cdot \left(\left(F \cdot C\right) \cdot \left(A \cdot A\right)\right)\right)}}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \]
                                3. Recombined 2 regimes into one program.
                                4. Final simplification7.7%

                                  \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -3.8 \cdot 10^{+155}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(-8 \cdot \left(\left(A \cdot A\right) \cdot \left(C \cdot F\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

                                Alternative 9: 8.7% 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 17.3%

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

                                  \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
                                3. Taylor expanded in A around -inf 7.5%

                                  \[\leadsto \frac{-\sqrt{\color{blue}{-16 \cdot \left({A}^{2} \cdot \left(C \cdot F\right)\right) + 4 \cdot \left(A \cdot \left(F \cdot {B}^{2}\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                4. Step-by-step derivation
                                  1. fma-def7.5%

                                    \[\leadsto \frac{-\sqrt{\color{blue}{\mathsf{fma}\left(-16, {A}^{2} \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(F \cdot {B}^{2}\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  2. unpow27.5%

                                    \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(-16, \color{blue}{\left(A \cdot A\right)} \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(F \cdot {B}^{2}\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  3. *-commutative7.5%

                                    \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(-16, \left(A \cdot A\right) \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \color{blue}{\left({B}^{2} \cdot F\right)}\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  4. unpow27.5%

                                    \[\leadsto \frac{-\sqrt{\mathsf{fma}\left(-16, \left(A \cdot A\right) \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(\color{blue}{\left(B \cdot B\right)} \cdot F\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                5. Simplified7.5%

                                  \[\leadsto \frac{-\sqrt{\color{blue}{\mathsf{fma}\left(-16, \left(A \cdot A\right) \cdot \left(C \cdot F\right), 4 \cdot \left(A \cdot \left(\left(B \cdot B\right) \cdot F\right)\right)\right)}}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                6. Taylor expanded in C around 0 3.6%

                                  \[\leadsto \color{blue}{-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)} \]
                                7. Step-by-step derivation
                                  1. associate-*r/3.6%

                                    \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]
                                  2. *-rgt-identity3.6%

                                    \[\leadsto -2 \cdot \frac{\color{blue}{\sqrt{A \cdot F}}}{B} \]
                                  3. *-commutative3.6%

                                    \[\leadsto -2 \cdot \frac{\sqrt{\color{blue}{F \cdot A}}}{B} \]
                                8. Simplified3.6%

                                  \[\leadsto \color{blue}{-2 \cdot \frac{\sqrt{F \cdot A}}{B}} \]
                                9. Final simplification3.6%

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

                                Alternative 10: 3.9% accurate, 6.1× speedup?

                                \[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -\sqrt{\frac{F}{A}} \end{array} \]
                                NOTE: B should be positive before calling this function
                                NOTE: A and C should be sorted in increasing order before calling this function.
                                (FPCore (A B C F) :precision binary64 (- (sqrt (/ F A))))
                                B = abs(B);
                                assert(A < C);
                                double code(double A, double B, double C, double F) {
                                	return -sqrt((F / A));
                                }
                                
                                NOTE: B should be positive before calling this function
                                NOTE: A and C should be sorted in increasing order before calling this function.
                                real(8) function code(a, b, c, f)
                                    real(8), intent (in) :: a
                                    real(8), intent (in) :: b
                                    real(8), intent (in) :: c
                                    real(8), intent (in) :: f
                                    code = -sqrt((f / a))
                                end function
                                
                                B = Math.abs(B);
                                assert A < C;
                                public static double code(double A, double B, double C, double F) {
                                	return -Math.sqrt((F / A));
                                }
                                
                                B = abs(B)
                                [A, C] = sort([A, C])
                                def code(A, B, C, F):
                                	return -math.sqrt((F / A))
                                
                                B = abs(B)
                                A, C = sort([A, C])
                                function code(A, B, C, F)
                                	return Float64(-sqrt(Float64(F / A)))
                                end
                                
                                B = abs(B)
                                A, C = num2cell(sort([A, C])){:}
                                function tmp = code(A, B, C, F)
                                	tmp = -sqrt((F / A));
                                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[Sqrt[N[(F / A), $MachinePrecision]], $MachinePrecision])
                                
                                \begin{array}{l}
                                B = |B|\\
                                [A, C] = \mathsf{sort}([A, C])\\
                                \\
                                -\sqrt{\frac{F}{A}}
                                \end{array}
                                
                                Derivation
                                1. Initial program 17.3%

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

                                  \[\leadsto \color{blue}{\frac{-\sqrt{F \cdot \left(\left(A - \left(\mathsf{hypot}\left(B, A - C\right) - C\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}} \]
                                3. Taylor expanded in A around -inf 7.3%

                                  \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\left(-0.5 \cdot \frac{{B}^{2}}{A} + -1 \cdot A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                4. Step-by-step derivation
                                  1. fma-def7.3%

                                    \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\mathsf{fma}\left(-0.5, \frac{{B}^{2}}{A}, -1 \cdot A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  2. unpow27.3%

                                    \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(-0.5, \frac{\color{blue}{B \cdot B}}{A}, -1 \cdot A\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                  3. mul-1-neg7.3%

                                    \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \mathsf{fma}\left(-0.5, \frac{B \cdot B}{A}, \color{blue}{-A}\right)\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                5. Simplified7.3%

                                  \[\leadsto \frac{-\sqrt{F \cdot \left(\left(A - \color{blue}{\mathsf{fma}\left(-0.5, \frac{B \cdot B}{A}, -A\right)}\right) \cdot \mathsf{fma}\left(C, A \cdot -8, 2 \cdot \left(B \cdot B\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)} \]
                                6. Taylor expanded in B around inf 2.7%

                                  \[\leadsto \color{blue}{-1 \cdot \sqrt{\frac{F}{A}}} \]
                                7. Step-by-step derivation
                                  1. mul-1-neg2.7%

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

                                  \[\leadsto \color{blue}{-\sqrt{\frac{F}{A}}} \]
                                9. Final simplification2.7%

                                  \[\leadsto -\sqrt{\frac{F}{A}} \]

                                Reproduce

                                ?
                                herbie shell --seed 2023224 
                                (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))))