ABCF->ab-angle b

Percentage Accurate: 18.7% → 49.5%

Time: 24.7s

Alternatives: 13

Speedup: 5.0×

Specification

Math

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

FPCore

(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)))

C

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;
}

Fortran

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

Java

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;
}

Python

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

Julia

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

MATLAB

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

Wolfram

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]]

Initial Program: 18.7% accurate, 1.0× speedup

Math

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

FPCore

(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)))

C

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;
}

Fortran

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

Java

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;
}

Python

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

Julia

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

MATLAB

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

Wolfram

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]]

Alternative 1: 49.5% accurate, 1.2× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := -\sqrt{2}\\ t_1 := \sqrt{F \cdot \left(C \cdot -4\right)} \cdot \frac{t_0}{\frac{C}{\sqrt{0.125}}}\\ t_2 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 1.05 \cdot 10^{-193}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 3.1 \cdot 10^{-78}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_2 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_2}\\ \mathbf{elif}\;B \leq 1.7 \cdot 10^{-53}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 7 \cdot 10^{+110}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(-4 \cdot A\right)\right)} \cdot \left(-\sqrt{\left(2 \cdot F\right) \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)}\right)}{\mathsf{fma}\left(B, B, \left(C \cdot -4\right) \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{t_0}{B}\\ \end{array} \end{array} \]

FPCore

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 (- (sqrt 2.0)))
        (t_1 (* (sqrt (* F (* C -4.0))) (/ t_0 (/ C (sqrt 0.125)))))
        (t_2 (- (* B B) (* 4.0 (* C A)))))
   (if (<= B 1.05e-193)
     t_1
     (if (<= B 3.1e-78)
       (/ (- (sqrt (* 2.0 (* t_2 (* F (* 2.0 A)))))) t_2)
       (if (<= B 1.7e-53)
         t_1
         (if (<= B 7e+110)
           (/
            (*
             (sqrt (fma B B (* C (* -4.0 A))))
             (- (sqrt (* (* 2.0 F) (+ A (- C (hypot B (- A C))))))))
            (fma B B (* (* C -4.0) A)))
           (* (sqrt (* F (- A (hypot A B)))) (/ t_0 B))))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = -sqrt(2.0);
	double t_1 = sqrt((F * (C * -4.0))) * (t_0 / (C / sqrt(0.125)));
	double t_2 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 1.05e-193) {
		tmp = t_1;
	} else if (B <= 3.1e-78) {
		tmp = -sqrt((2.0 * (t_2 * (F * (2.0 * A))))) / t_2;
	} else if (B <= 1.7e-53) {
		tmp = t_1;
	} else if (B <= 7e+110) {
		tmp = (sqrt(fma(B, B, (C * (-4.0 * A)))) * -sqrt(((2.0 * F) * (A + (C - hypot(B, (A - C))))))) / fma(B, B, ((C * -4.0) * A));
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (t_0 / B);
	}
	return tmp;
}

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(-sqrt(2.0))
	t_1 = Float64(sqrt(Float64(F * Float64(C * -4.0))) * Float64(t_0 / Float64(C / sqrt(0.125))))
	t_2 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 1.05e-193)
		tmp = t_1;
	elseif (B <= 3.1e-78)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_2 * Float64(F * Float64(2.0 * A)))))) / t_2);
	elseif (B <= 1.7e-53)
		tmp = t_1;
	elseif (B <= 7e+110)
		tmp = Float64(Float64(sqrt(fma(B, B, Float64(C * Float64(-4.0 * A)))) * Float64(-sqrt(Float64(Float64(2.0 * F) * Float64(A + Float64(C - hypot(B, Float64(A - C)))))))) / fma(B, B, Float64(Float64(C * -4.0) * A)));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(t_0 / B));
	end
	return tmp
end

Wolfram

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[Sqrt[2.0], $MachinePrecision])}, Block[{t$95$1 = N[(N[Sqrt[N[(F * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 / N[(C / N[Sqrt[0.125], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.05e-193], t$95$1, If[LessEqual[B, 3.1e-78], N[((-N[Sqrt[N[(2.0 * N[(t$95$2 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$2), $MachinePrecision], If[LessEqual[B, 1.7e-53], t$95$1, If[LessEqual[B, 7e+110], N[(N[(N[Sqrt[N[(B * B + N[(C * N[(-4.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[N[(N[(2.0 * F), $MachinePrecision] * N[(A + N[(C - N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / N[(B * B + N[(N[(C * -4.0), $MachinePrecision] * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 / B), $MachinePrecision]), $MachinePrecision]]]]]]]]

Derivation

1. Split input into 4 regimes

2. if B < 1.05e-193 or 3.10000000000000018e-78 < B < 1.7e-53

   1. Initial program 22.9%

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

   3. Simplified 22.9%

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

      1. flip-- 9.1%

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

      2. add-sqr-sqrt 9.1%

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

      3. fma-def 9.1%

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

      4. unpow2 9.1%

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

      5. hypot-udef 9.1%

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

   5. Applied egg-rr 9.1%

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

   6. Taylor expanded in A around -inf 18.2%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2} \cdot \sqrt{0.125}}{C} \cdot \sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}\right)} \]
   7. Step-by-step derivation

      1. mul-1-neg 18.2%

         \[\leadsto \color{blue}{-\frac{\sqrt{2} \cdot \sqrt{0.125}}{C} \cdot \sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}} \]

      2. distribute-rgt-neg-in 18.2%

         \[\leadsto \color{blue}{\frac{\sqrt{2} \cdot \sqrt{0.125}}{C} \cdot \left(-\sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}\right)} \]

      3. associate-/l* 18.3%

         \[\leadsto \color{blue}{\frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}}} \cdot \left(-\sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}\right) \]

      4. *-commutative 18.3%

         \[\leadsto \frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{\color{blue}{F \cdot \left(-2 \cdot C - 2 \cdot C\right)}}\right) \]

      5. distribute-rgt-out-- 18.3%

         \[\leadsto \frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{F \cdot \color{blue}{\left(C \cdot \left(-2 - 2\right)\right)}}\right) \]

      6. metadata-eval 18.3%

         \[\leadsto \frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{F \cdot \left(C \cdot \color{blue}{-4}\right)}\right) \]

   8. Simplified 18.3%

      \[\leadsto \color{blue}{\frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{F \cdot \left(C \cdot -4\right)}\right)} \]

   if 1.05e-193 < B < 3.10000000000000018e-78

   1. Initial program 29.1%

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

   3. Simplified 29.1%

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

   4. Taylor expanded in A around -inf 33.7%

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

      1. *-commutative 33.7%

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

   6. Simplified 33.7%

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

      1. distribute-frac-neg 33.7%

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

      2. associate-*l* 36.4%

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

   8. Applied egg-rr 36.4%

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

   if 1.7e-53 < B < 6.9999999999999998e110

   1. Initial program 33.8%

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

   3. Simplified 40.6%

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

      1. sqrt-prod 46.2%

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

      2. associate-*r* 46.2%

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

      3. *-commutative 46.2%

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

      4. associate-*l* 46.3%

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

      5. associate--r- 45.3%

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

      6. +-commutative 45.3%

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

   5. Applied egg-rr 45.3%

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

      1. associate-*r* 45.3%

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

   7. Simplified 45.3%

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

   if 6.9999999999999998e110 < B

   1. Initial program 0.6%

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

   3. Simplified 0.6%

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

   4. Taylor expanded in C around 0 8.0%

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

      1. mul-1-neg 8.0%

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

      2. *-commutative 8.0%

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

      3. +-commutative 8.0%

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

      4. unpow2 8.0%

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

      5. unpow2 8.0%

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

      6. hypot-def 43.1%

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

   6. Simplified 43.1%

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

4. Final simplification 29.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.05 \cdot 10^{-193}:\\ \;\;\;\;\sqrt{F \cdot \left(C \cdot -4\right)} \cdot \frac{-\sqrt{2}}{\frac{C}{\sqrt{0.125}}}\\ \mathbf{elif}\;B \leq 3.1 \cdot 10^{-78}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 1.7 \cdot 10^{-53}:\\ \;\;\;\;\sqrt{F \cdot \left(C \cdot -4\right)} \cdot \frac{-\sqrt{2}}{\frac{C}{\sqrt{0.125}}}\\ \mathbf{elif}\;B \leq 7 \cdot 10^{+110}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(-4 \cdot A\right)\right)} \cdot \left(-\sqrt{\left(2 \cdot F\right) \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)}\right)}{\mathsf{fma}\left(B, B, \left(C \cdot -4\right) \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 2: 49.0% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := -\sqrt{2}\\ t_1 := \sqrt{F \cdot \left(C \cdot -4\right)} \cdot \frac{t_0}{\frac{C}{\sqrt{0.125}}}\\ t_2 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 7.8 \cdot 10^{-194}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;B \leq 2.2 \cdot 10^{-78}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_2 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_2}\\ \mathbf{elif}\;B \leq 3.7 \cdot 10^{-15}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{t_0}{B}\\ \end{array} \end{array} \]

FPCore

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 (- (sqrt 2.0)))
        (t_1 (* (sqrt (* F (* C -4.0))) (/ t_0 (/ C (sqrt 0.125)))))
        (t_2 (- (* B B) (* 4.0 (* C A)))))
   (if (<= B 7.8e-194)
     t_1
     (if (<= B 2.2e-78)
       (/ (- (sqrt (* 2.0 (* t_2 (* F (* 2.0 A)))))) t_2)
       (if (<= B 3.7e-15) t_1 (* (sqrt (* F (- A (hypot A B)))) (/ t_0 B)))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = -sqrt(2.0);
	double t_1 = sqrt((F * (C * -4.0))) * (t_0 / (C / sqrt(0.125)));
	double t_2 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 7.8e-194) {
		tmp = t_1;
	} else if (B <= 2.2e-78) {
		tmp = -sqrt((2.0 * (t_2 * (F * (2.0 * A))))) / t_2;
	} else if (B <= 3.7e-15) {
		tmp = t_1;
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (t_0 / B);
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = -Math.sqrt(2.0);
	double t_1 = Math.sqrt((F * (C * -4.0))) * (t_0 / (C / Math.sqrt(0.125)));
	double t_2 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 7.8e-194) {
		tmp = t_1;
	} else if (B <= 2.2e-78) {
		tmp = -Math.sqrt((2.0 * (t_2 * (F * (2.0 * A))))) / t_2;
	} else if (B <= 3.7e-15) {
		tmp = t_1;
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(A, B)))) * (t_0 / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = -math.sqrt(2.0)
	t_1 = math.sqrt((F * (C * -4.0))) * (t_0 / (C / math.sqrt(0.125)))
	t_2 = (B * B) - (4.0 * (C * A))
	tmp = 0
	if B <= 7.8e-194:
		tmp = t_1
	elif B <= 2.2e-78:
		tmp = -math.sqrt((2.0 * (t_2 * (F * (2.0 * A))))) / t_2
	elif B <= 3.7e-15:
		tmp = t_1
	else:
		tmp = math.sqrt((F * (A - math.hypot(A, B)))) * (t_0 / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(-sqrt(2.0))
	t_1 = Float64(sqrt(Float64(F * Float64(C * -4.0))) * Float64(t_0 / Float64(C / sqrt(0.125))))
	t_2 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 7.8e-194)
		tmp = t_1;
	elseif (B <= 2.2e-78)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_2 * Float64(F * Float64(2.0 * A)))))) / t_2);
	elseif (B <= 3.7e-15)
		tmp = t_1;
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(t_0 / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = -sqrt(2.0);
	t_1 = sqrt((F * (C * -4.0))) * (t_0 / (C / sqrt(0.125)));
	t_2 = (B * B) - (4.0 * (C * A));
	tmp = 0.0;
	if (B <= 7.8e-194)
		tmp = t_1;
	elseif (B <= 2.2e-78)
		tmp = -sqrt((2.0 * (t_2 * (F * (2.0 * A))))) / t_2;
	elseif (B <= 3.7e-15)
		tmp = t_1;
	else
		tmp = sqrt((F * (A - hypot(A, B)))) * (t_0 / B);
	end
	tmp_2 = tmp;
end

Wolfram

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[Sqrt[2.0], $MachinePrecision])}, Block[{t$95$1 = N[(N[Sqrt[N[(F * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 / N[(C / N[Sqrt[0.125], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 7.8e-194], t$95$1, If[LessEqual[B, 2.2e-78], N[((-N[Sqrt[N[(2.0 * N[(t$95$2 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$2), $MachinePrecision], If[LessEqual[B, 3.7e-15], t$95$1, N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(t$95$0 / B), $MachinePrecision]), $MachinePrecision]]]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 7.7999999999999997e-194 or 2.1999999999999999e-78 < B < 3.70000000000000017e-15

   1. Initial program 24.5%

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

   3. Simplified 24.5%

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

      1. flip-- 9.7%

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

      2. add-sqr-sqrt 9.7%

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

      3. fma-def 9.7%

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

      4. unpow2 9.7%

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

      5. hypot-udef 9.7%

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

   5. Applied egg-rr 9.7%

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

   6. Taylor expanded in A around -inf 17.6%

      \[\leadsto \color{blue}{-1 \cdot \left(\frac{\sqrt{2} \cdot \sqrt{0.125}}{C} \cdot \sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}\right)} \]
   7. Step-by-step derivation

      1. mul-1-neg 17.6%

         \[\leadsto \color{blue}{-\frac{\sqrt{2} \cdot \sqrt{0.125}}{C} \cdot \sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}} \]

      2. distribute-rgt-neg-in 17.6%

         \[\leadsto \color{blue}{\frac{\sqrt{2} \cdot \sqrt{0.125}}{C} \cdot \left(-\sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}\right)} \]

      3. associate-/l* 17.7%

         \[\leadsto \color{blue}{\frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}}} \cdot \left(-\sqrt{\left(-2 \cdot C - 2 \cdot C\right) \cdot F}\right) \]

      4. *-commutative 17.7%

         \[\leadsto \frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{\color{blue}{F \cdot \left(-2 \cdot C - 2 \cdot C\right)}}\right) \]

      5. distribute-rgt-out-- 17.7%

         \[\leadsto \frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{F \cdot \color{blue}{\left(C \cdot \left(-2 - 2\right)\right)}}\right) \]

      6. metadata-eval 17.7%

         \[\leadsto \frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{F \cdot \left(C \cdot \color{blue}{-4}\right)}\right) \]

   8. Simplified 17.7%

      \[\leadsto \color{blue}{\frac{\sqrt{2}}{\frac{C}{\sqrt{0.125}}} \cdot \left(-\sqrt{F \cdot \left(C \cdot -4\right)}\right)} \]

   if 7.7999999999999997e-194 < B < 2.1999999999999999e-78

   1. Initial program 29.1%

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

   3. Simplified 29.1%

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

   4. Taylor expanded in A around -inf 33.7%

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

      1. *-commutative 33.7%

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

   6. Simplified 33.7%

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

      1. distribute-frac-neg 33.7%

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

      2. associate-*l* 36.4%

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

   8. Applied egg-rr 36.4%

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

   if 3.70000000000000017e-15 < B

   1. Initial program 10.2%

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

   3. Simplified 10.2%

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

   4. Taylor expanded in C around 0 15.1%

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

      1. mul-1-neg 15.1%

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

      2. *-commutative 15.1%

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

      3. +-commutative 15.1%

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

      4. unpow2 15.1%

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

      5. unpow2 15.1%

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

      6. hypot-def 39.0%

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

   6. Simplified 39.0%

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

4. Final simplification 26.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 7.8 \cdot 10^{-194}:\\ \;\;\;\;\sqrt{F \cdot \left(C \cdot -4\right)} \cdot \frac{-\sqrt{2}}{\frac{C}{\sqrt{0.125}}}\\ \mathbf{elif}\;B \leq 2.2 \cdot 10^{-78}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 3.7 \cdot 10^{-15}:\\ \;\;\;\;\sqrt{F \cdot \left(C \cdot -4\right)} \cdot \frac{-\sqrt{2}}{\frac{C}{\sqrt{0.125}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 3: 46.3% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 6.6 \cdot 10^{-12}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

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 (* C A)))))
   (if (<= B 6.6e-12)
     (- (/ (sqrt (* 2.0 (* t_0 (* F (* 2.0 A))))) t_0))
     (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 6.6e-12) {
		tmp = -(sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Java

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 * (C * A));
	double tmp;
	if (B <= 6.6e-12) {
		tmp = -(Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(A, B)))) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	tmp = 0
	if B <= 6.6e-12:
		tmp = -(math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0)
	else:
		tmp = math.sqrt((F * (A - math.hypot(A, B)))) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 6.6e-12)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A))))) / t_0));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	tmp = 0.0;
	if (B <= 6.6e-12)
		tmp = -(sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
	else
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 6.6e-12], (-N[(N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 6.6000000000000001e-12

   1. Initial program 25.4%

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

   3. Simplified 25.4%

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

   4. Taylor expanded in A around -inf 17.8%

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

      1. *-commutative 17.8%

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

   6. Simplified 17.8%

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

      1. distribute-frac-neg 17.8%

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

      2. associate-*l* 19.4%

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

   8. Applied egg-rr 19.4%

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

   if 6.6000000000000001e-12 < B

   1. Initial program 9.1%

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

   3. Simplified 9.1%

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

   4. Taylor expanded in C around 0 15.1%

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

      1. mul-1-neg 15.1%

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

      2. *-commutative 15.1%

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

      3. +-commutative 15.1%

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

      4. unpow2 15.1%

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

      5. unpow2 15.1%

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

      6. hypot-def 39.3%

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

   6. Simplified 39.3%

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

4. Final simplification 25.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6.6 \cdot 10^{-12}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 4: 44.5% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 3.1 \cdot 10^{-55}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 2.2 \cdot 10^{+84}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]

FPCore

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 (* C A)))) (t_1 (- (* B B) (* 4.0 (* C A)))))
   (if (<= B 3.1e-55)
     (- (/ (sqrt (* 2.0 (* t_1 (* F (* 2.0 A))))) t_1))
     (if (<= B 2.2e+84)
       (/ (- (sqrt (* 2.0 (* t_0 (* F (+ A (- C (hypot B (- A C))))))))) t_0)
       (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A B)))))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (C * A));
	double t_1 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 3.1e-55) {
		tmp = -(sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1);
	} else if (B <= 2.2e+84) {
		tmp = -sqrt((2.0 * (t_0 * (F * (A + (C - hypot(B, (A - C)))))))) / t_0;
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	}
	return tmp;
}

Java

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 * (C * A));
	double t_1 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 3.1e-55) {
		tmp = -(Math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1);
	} else if (B <= 2.2e+84) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (A + (C - Math.hypot(B, (A - C)))))))) / t_0;
	} else {
		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt((F * (A - B)));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) + (-4.0 * (C * A))
	t_1 = (B * B) - (4.0 * (C * A))
	tmp = 0
	if B <= 3.1e-55:
		tmp = -(math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1)
	elif B <= 2.2e+84:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (A + (C - math.hypot(B, (A - C)))))))) / t_0
	else:
		tmp = (math.sqrt(2.0) / B) * -math.sqrt((F * (A - B)))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	t_1 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 3.1e-55)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(2.0 * A))))) / t_1));
	elseif (B <= 2.2e+84)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(A + Float64(C - hypot(B, Float64(A - C))))))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(F * Float64(A - B)))));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) + (-4.0 * (C * A));
	t_1 = (B * B) - (4.0 * (C * A));
	tmp = 0.0;
	if (B <= 3.1e-55)
		tmp = -(sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1);
	elseif (B <= 2.2e+84)
		tmp = -sqrt((2.0 * (t_0 * (F * (A + (C - hypot(B, (A - C)))))))) / t_0;
	else
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	end
	tmp_2 = tmp;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.1e-55], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$1), $MachinePrecision]), If[LessEqual[B, 2.2e+84], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(A + N[(C - N[Sqrt[B ^ 2 + N[(A - C), $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $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]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 3.09999999999999997e-55

   1. Initial program 23.9%

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

   3. Simplified 23.9%

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

   4. Taylor expanded in A around -inf 19.3%

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

      1. *-commutative 19.3%

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

   6. Simplified 19.3%

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

      1. distribute-frac-neg 19.3%

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

      2. associate-*l* 21.0%

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

   8. Applied egg-rr 21.0%

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

   if 3.09999999999999997e-55 < B < 2.1999999999999998e84

   1. Initial program 33.8%

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

   3. Simplified 33.8%

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

      1. distribute-frac-neg 33.8%

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

   5. Applied egg-rr 39.8%

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

   if 2.1999999999999998e84 < B

   1. Initial program 2.3%

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

   3. Simplified 2.3%

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

   4. Taylor expanded in C around 0 10.6%

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

      1. mul-1-neg 10.6%

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

      2. *-commutative 10.6%

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

      3. +-commutative 10.6%

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

      4. unpow2 10.6%

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

      5. unpow2 10.6%

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

      6. hypot-def 42.9%

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

   6. Simplified 42.9%

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

   7. Taylor expanded in A around 0 41.4%

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

      1. neg-mul-1 41.4%

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

      2. unsub-neg 41.4%

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

   9. Simplified 41.4%

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

4. Final simplification 28.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.1 \cdot 10^{-55}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 2.2 \cdot 10^{+84}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(B, A - C\right)\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

Alternative 5: 43.7% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;B \leq 3.6 \cdot 10^{-54}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 6.5 \cdot 10^{+23}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(A, B\right)\right) \cdot t_1\right)}}{t_0}\\ \mathbf{elif}\;B \leq 1.58 \cdot 10^{+38}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(A + \left(A - -0.5 \cdot \frac{\left(A \cdot A - A \cdot A\right) - B \cdot B}{C}\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

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 (* C A)))) (t_1 (* F t_0)))
   (if (<= B 3.6e-54)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (if (<= B 6.5e+23)
       (/ (- (sqrt (* 2.0 (* (- A (hypot A B)) t_1)))) t_0)
       (if (<= B 1.58e+38)
         (-
          (/
           (sqrt
            (*
             2.0
             (*
              t_1
              (+ A (- A (* -0.5 (/ (- (- (* A A) (* A A)) (* B B)) C)))))))
           t_0))
         (* (sqrt (* F (- A B))) (/ (- (sqrt 2.0)) B)))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = F * t_0;
	double tmp;
	if (B <= 3.6e-54) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else if (B <= 6.5e+23) {
		tmp = -sqrt((2.0 * ((A - hypot(A, B)) * t_1))) / t_0;
	} else if (B <= 1.58e+38) {
		tmp = -(sqrt((2.0 * (t_1 * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0);
	} else {
		tmp = sqrt((F * (A - B))) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Java

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 * (C * A));
	double t_1 = F * t_0;
	double tmp;
	if (B <= 3.6e-54) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else if (B <= 6.5e+23) {
		tmp = -Math.sqrt((2.0 * ((A - Math.hypot(A, B)) * t_1))) / t_0;
	} else if (B <= 1.58e+38) {
		tmp = -(Math.sqrt((2.0 * (t_1 * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0);
	} else {
		tmp = Math.sqrt((F * (A - B))) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = F * t_0
	tmp = 0
	if B <= 3.6e-54:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	elif B <= 6.5e+23:
		tmp = -math.sqrt((2.0 * ((A - math.hypot(A, B)) * t_1))) / t_0
	elif B <= 1.58e+38:
		tmp = -(math.sqrt((2.0 * (t_1 * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0)
	else:
		tmp = math.sqrt((F * (A - B))) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(F * t_0)
	tmp = 0.0
	if (B <= 3.6e-54)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	elseif (B <= 6.5e+23)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(A - hypot(A, B)) * t_1)))) / t_0);
	elseif (B <= 1.58e+38)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(A + Float64(A - Float64(-0.5 * Float64(Float64(Float64(Float64(A * A) - Float64(A * A)) - Float64(B * B)) / C))))))) / t_0));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - B))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = F * t_0;
	tmp = 0.0;
	if (B <= 3.6e-54)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	elseif (B <= 6.5e+23)
		tmp = -sqrt((2.0 * ((A - hypot(A, B)) * t_1))) / t_0;
	elseif (B <= 1.58e+38)
		tmp = -(sqrt((2.0 * (t_1 * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0);
	else
		tmp = sqrt((F * (A - B))) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, If[LessEqual[B, 3.6e-54], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 6.5e+23], N[((-N[Sqrt[N[(2.0 * N[(N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 1.58e+38], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + N[(A - N[(-0.5 * N[(N[(N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision] - N[(B * B), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[(N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 4 regimes

2. if B < 3.59999999999999976e-54

   1. Initial program 23.8%

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

   3. Simplified 23.8%

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

   4. Taylor expanded in A around -inf 19.2%

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

      1. *-commutative 19.2%

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

   6. Simplified 19.2%

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

      1. distribute-frac-neg 19.2%

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

      2. associate-*l* 20.9%

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

   8. Applied egg-rr 20.9%

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

   if 3.59999999999999976e-54 < B < 6.4999999999999996e23

   1. Initial program 42.4%

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

   3. Simplified 42.4%

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

   4. Taylor expanded in C around 0 22.5%

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

      1. +-commutative 22.5%

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

      2. unpow2 22.5%

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

      3. unpow2 22.5%

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

      4. hypot-def 22.8%

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

   6. Simplified 22.8%

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

   if 6.4999999999999996e23 < B < 1.58e38

   1. Initial program 17.2%

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

   3. Simplified 17.2%

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

   4. Taylor expanded in C around inf 50.3%

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

      1. associate--l+ 50.3%

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

      2. associate--l+ 50.3%

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

      3. unpow2 50.3%

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

      4. unpow2 50.3%

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

      5. unpow2 50.3%

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

      6. mul-1-neg 50.3%

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

      7. mul-1-neg 50.3%

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

      8. sqr-neg 50.3%

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

      9. mul-1-neg 50.3%

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

   6. Simplified 50.3%

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

   if 1.58e38 < B

   1. Initial program 5.1%

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

   3. Simplified 5.1%

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

   4. Taylor expanded in C around 0 14.1%

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

      1. mul-1-neg 14.1%

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

      2. *-commutative 14.1%

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

      3. +-commutative 14.1%

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

      4. unpow2 14.1%

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

      5. unpow2 14.1%

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

      6. hypot-def 42.3%

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

   6. Simplified 42.3%

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

   7. Taylor expanded in A around 0 41.0%

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

      1. neg-mul-1 41.0%

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

      2. unsub-neg 41.0%

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

   9. Simplified 41.0%

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

4. Final simplification 26.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.6 \cdot 10^{-54}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 6.5 \cdot 10^{+23}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A - \mathsf{hypot}\left(A, B\right)\right) \cdot \left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 1.58 \cdot 10^{+38}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(A + \left(A - -0.5 \cdot \frac{\left(A \cdot A - A \cdot A\right) - B \cdot B}{C}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 6: 42.8% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 6 \cdot 10^{+39}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]

FPCore

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 (* C A)))))
   (if (<= B 6e+39)
     (- (/ (sqrt (* 2.0 (* t_0 (* F (* 2.0 A))))) t_0))
     (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A B))))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 6e+39) {
		tmp = -(sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	}
	return tmp;
}

Fortran

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 * (c * a))
    if (b <= 6d+39) then
        tmp = -(sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / t_0)
    else
        tmp = (sqrt(2.0d0) / b) * -sqrt((f * (a - b)))
    end if
    code = tmp
end function

Java

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 * (C * A));
	double tmp;
	if (B <= 6e+39) {
		tmp = -(Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
	} else {
		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt((F * (A - B)));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	tmp = 0
	if B <= 6e+39:
		tmp = -(math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0)
	else:
		tmp = (math.sqrt(2.0) / B) * -math.sqrt((F * (A - B)))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 6e+39)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_0 * Float64(F * 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

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	tmp = 0.0;
	if (B <= 6e+39)
		tmp = -(sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
	else
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	end
	tmp_2 = tmp;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 6e+39], (-N[(N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $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]]]

Derivation

1. Split input into 2 regimes

2. if B < 5.9999999999999999e39

   1. Initial program 25.8%

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

   3. Simplified 25.8%

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

   4. Taylor expanded in A around -inf 17.4%

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

      1. *-commutative 17.4%

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

   6. Simplified 17.4%

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

      1. distribute-frac-neg 17.4%

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

      2. associate-*l* 18.8%

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

   8. Applied egg-rr 18.8%

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

   if 5.9999999999999999e39 < B

   1. Initial program 5.1%

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

   3. Simplified 5.1%

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

   4. Taylor expanded in C around 0 14.1%

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

      1. mul-1-neg 14.1%

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

      2. *-commutative 14.1%

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

      3. +-commutative 14.1%

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

      4. unpow2 14.1%

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

      5. unpow2 14.1%

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

      6. hypot-def 42.3%

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

   6. Simplified 42.3%

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

   7. Taylor expanded in A around 0 41.0%

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

      1. neg-mul-1 41.0%

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

      2. unsub-neg 41.0%

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

   9. Simplified 41.0%

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

4. Final simplification 24.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6 \cdot 10^{+39}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

Alternative 7: 42.4% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 1.5 \cdot 10^{+38}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{-B \cdot F} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

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 (* C A)))))
   (if (<= B 1.5e+38)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (* (sqrt (- (* B F))) (/ (- (sqrt 2.0)) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (B <= 1.5e+38) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = sqrt(-(B * F)) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Fortran

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 * (c * a))
    if (b <= 1.5d+38) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / t_0
    else
        tmp = sqrt(-(b * f)) * (-sqrt(2.0d0) / b)
    end if
    code = tmp
end function

Java

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 * (C * A));
	double tmp;
	if (B <= 1.5e+38) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = Math.sqrt(-(B * F)) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	tmp = 0
	if B <= 1.5e+38:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	else:
		tmp = math.sqrt(-(B * F)) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 1.5e+38)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(-Float64(B * F))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	tmp = 0.0;
	if (B <= 1.5e+38)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	else
		tmp = sqrt(-(B * F)) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.5e+38], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[(-N[(B * F), $MachinePrecision])], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 1.5000000000000001e38

   1. Initial program 25.8%

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

   3. Simplified 25.8%

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

   4. Taylor expanded in A around -inf 17.4%

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

      1. *-commutative 17.4%

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

   6. Simplified 17.4%

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

      1. distribute-frac-neg 17.4%

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

      2. associate-*l* 18.8%

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

   8. Applied egg-rr 18.8%

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

   if 1.5000000000000001e38 < B

   1. Initial program 5.1%

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

   3. Simplified 5.1%

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

   4. Taylor expanded in C around 0 14.1%

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

      1. mul-1-neg 14.1%

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

      2. *-commutative 14.1%

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

      3. +-commutative 14.1%

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

      4. unpow2 14.1%

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

      5. unpow2 14.1%

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

      6. hypot-def 42.3%

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

   6. Simplified 42.3%

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

   7. Taylor expanded in A around 0 42.0%

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

      1. neg-mul-1 42.0%

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

   9. Simplified 42.0%

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

4. Final simplification 25.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.5 \cdot 10^{+38}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{-B \cdot F} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 8: 29.7% accurate, 4.3× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;A \leq -2.2 \cdot 10^{-97}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + \left(A - -0.5 \cdot \frac{\left(A \cdot A - A \cdot A\right) - B \cdot B}{C}\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

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 (* C A)))))
   (if (<= A -2.2e-97)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (-
      (/
       (sqrt
        (*
         2.0
         (*
          (* F t_0)
          (+ A (- A (* -0.5 (/ (- (- (* A A) (* A A)) (* B B)) C)))))))
       t_0)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (A <= -2.2e-97) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = -(sqrt((2.0 * ((F * t_0) * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0);
	}
	return tmp;
}

Fortran

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 * (c * a))
    if (a <= (-2.2d-97)) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / t_0
    else
        tmp = -(sqrt((2.0d0 * ((f * t_0) * (a + (a - ((-0.5d0) * ((((a * a) - (a * a)) - (b * b)) / c))))))) / t_0)
    end if
    code = tmp
end function

Java

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 * (C * A));
	double tmp;
	if (A <= -2.2e-97) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = -(Math.sqrt((2.0 * ((F * t_0) * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	tmp = 0
	if A <= -2.2e-97:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	else:
		tmp = -(math.sqrt((2.0 * ((F * t_0) * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if (A <= -2.2e-97)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + Float64(A - Float64(-0.5 * Float64(Float64(Float64(Float64(A * A) - Float64(A * A)) - Float64(B * B)) / C))))))) / t_0));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	tmp = 0.0;
	if (A <= -2.2e-97)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	else
		tmp = -(sqrt((2.0 * ((F * t_0) * (A + (A - (-0.5 * ((((A * A) - (A * A)) - (B * B)) / C))))))) / t_0);
	end
	tmp_2 = tmp;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[A, -2.2e-97], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + N[(A - N[(-0.5 * N[(N[(N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision] - N[(B * B), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision])]]

Derivation

1. Split input into 2 regimes

2. if A < -2.1999999999999999e-97

   1. Initial program 22.6%

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

   3. Simplified 22.6%

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

   4. Taylor expanded in A around -inf 28.9%

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

      1. *-commutative 28.9%

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

   6. Simplified 28.9%

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

      1. distribute-frac-neg 28.9%

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

      2. associate-*l* 32.4%

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

   8. Applied egg-rr 32.4%

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

   if -2.1999999999999999e-97 < A

   1. Initial program 19.2%

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

   3. Simplified 19.2%

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

   4. Taylor expanded in C around inf 7.4%

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

      1. associate--l+ 7.4%

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

      2. associate--l+ 7.5%

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

      3. unpow2 7.5%

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

      4. unpow2 7.5%

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

      5. unpow2 7.5%

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

      6. mul-1-neg 7.5%

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

      7. mul-1-neg 7.5%

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

      8. sqr-neg 7.5%

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

      9. mul-1-neg 7.5%

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

   6. Simplified 7.5%

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

4. Final simplification 15.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -2.2 \cdot 10^{-97}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(A + \left(A - -0.5 \cdot \frac{\left(A \cdot A - A \cdot A\right) - B \cdot B}{C}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \end{array} \]

Alternative 9: 26.1% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \frac{-{\left(2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)\right)}^{0.5}}{t_0} \end{array} \end{array} \]

FPCore

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 (* C A)))))
   (/ (- (pow (* 2.0 (* t_0 (* F (* 2.0 A)))) 0.5)) t_0)))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	return -pow((2.0 * (t_0 * (F * (2.0 * A)))), 0.5) / t_0;
}

Fortran

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 * (c * a))
    code = -((2.0d0 * (t_0 * (f * (2.0d0 * a)))) ** 0.5d0) / t_0
end function

Java

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 * (C * A));
	return -Math.pow((2.0 * (t_0 * (F * (2.0 * A)))), 0.5) / t_0;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	return -math.pow((2.0 * (t_0 * (F * (2.0 * A)))), 0.5) / t_0

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	return Float64(Float64(-(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))) ^ 0.5)) / t_0)
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	tmp = -((2.0 * (t_0 * (F * (2.0 * A)))) ^ 0.5) / t_0;
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[((-N[Power[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 0.5], $MachinePrecision]) / t$95$0), $MachinePrecision]]

Derivation

1. Initial program 20.3%

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

3. Simplified 20.3%

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

4. Taylor expanded in A around -inf 12.8%

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

   1. *-commutative 12.8%

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

6. Simplified 12.8%

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

   1. pow1/2 13.0%

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

   2. associate-*l* 14.1%

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

8. Applied egg-rr 14.1%

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

9. Final simplification 14.1%

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

Alternative 10: 26.1% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ -\frac{\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0} \end{array} \end{array} \]

FPCore

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 (* C A)))))
   (- (/ (sqrt (* 2.0 (* t_0 (* F (* 2.0 A))))) t_0))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	return -(sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
}

Fortran

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 * (c * a))
    code = -(sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / t_0)
end function

Java

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 * (C * A));
	return -(Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	return -(math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0)

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	return Float64(-Float64(sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A))))) / t_0))
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	tmp = -(sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0);
end

Wolfram

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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, (-N[(N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision])]

Derivation

1. Initial program 20.3%

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

3. Simplified 20.3%

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

4. Taylor expanded in A around -inf 12.8%

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

   1. *-commutative 12.8%

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

6. Simplified 12.8%

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

   1. distribute-frac-neg 12.8%

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

   2. associate-*l* 13.9%

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

8. Applied egg-rr 13.9%

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

9. Final simplification 13.9%

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

Alternative 11: 22.7% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 1.85 \cdot 10^{-15}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{F \cdot A}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (if (<= B 1.85e-15)
   (/
    (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* A (* C F)))))))
    (- (* B B) (* 4.0 (* C A))))
   (* -2.0 (/ (sqrt (* F A)) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 1.85e-15) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (sqrt((F * A)) / B);
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: tmp
    if (b <= 1.85d-15) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (a * (c * f)))))) / ((b * b) - (4.0d0 * (c * a)))
    else
        tmp = (-2.0d0) * (sqrt((f * a)) / b)
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 1.85e-15) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (Math.sqrt((F * A)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 1.85e-15:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)))
	else:
		tmp = -2.0 * (math.sqrt((F * A)) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 1.85e-15)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(-4.0 * Float64(A * Float64(C * F))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(C * A))));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(F * A)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 1.85e-15)
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)));
	else
		tmp = -2.0 * (sqrt((F * A)) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := If[LessEqual[B, 1.85e-15], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(-4.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(F * A), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 1.85000000000000008e-15

   1. Initial program 25.0%

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

   3. Simplified 25.0%

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

   4. Taylor expanded in A around -inf 17.9%

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

      1. *-commutative 17.9%

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

   6. Simplified 17.9%

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

   7. Taylor expanded in B around 0 12.9%

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

   if 1.85000000000000008e-15 < B

   1. Initial program 10.2%

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

   3. Simplified 10.2%

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

   4. Taylor expanded in A around -inf 2.1%

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

      1. *-commutative 2.1%

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

   6. Simplified 2.1%

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

   7. Taylor expanded in B around inf 2.4%

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

      1. associate-*r/ 2.4%

         \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]

      2. *-rgt-identity 2.4%

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

   9. Simplified 2.4%

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

4. Final simplification 9.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.85 \cdot 10^{-15}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{F \cdot A}}{B}\\ \end{array} \]

Alternative 12: 26.3% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 4.5 \cdot 10^{-11}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(\left(C \cdot -4\right) \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{F \cdot A}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (if (<= B 4.5e-11)
   (/
    (- (sqrt (* 2.0 (* (* 2.0 A) (* F (* (* C -4.0) A))))))
    (- (* B B) (* 4.0 (* C A))))
   (* -2.0 (/ (sqrt (* F A)) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 4.5e-11) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * ((C * -4.0) * A))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (sqrt((F * A)) / B);
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: tmp
    if (b <= 4.5d-11) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * (f * ((c * (-4.0d0)) * a))))) / ((b * b) - (4.0d0 * (c * a)))
    else
        tmp = (-2.0d0) * (sqrt((f * a)) / b)
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 4.5e-11) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (F * ((C * -4.0) * A))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (Math.sqrt((F * A)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 4.5e-11:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (F * ((C * -4.0) * A))))) / ((B * B) - (4.0 * (C * A)))
	else:
		tmp = -2.0 * (math.sqrt((F * A)) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 4.5e-11)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(F * Float64(Float64(C * -4.0) * A)))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(C * A))));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(F * A)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 4.5e-11)
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * ((C * -4.0) * A))))) / ((B * B) - (4.0 * (C * A)));
	else
		tmp = -2.0 * (sqrt((F * A)) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := If[LessEqual[B, 4.5e-11], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(F * N[(N[(C * -4.0), $MachinePrecision] * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(F * A), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 4.5e-11

   1. Initial program 25.4%

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

   3. Simplified 25.4%

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

   4. Taylor expanded in A around -inf 17.8%

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

      1. *-commutative 17.8%

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

   6. Simplified 17.8%

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

   7. Taylor expanded in B around 0 14.9%

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

      1. *-commutative 14.9%

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

      2. associate-*r* 14.9%

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

      3. *-commutative 14.9%

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

   9. Simplified 14.9%

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

   if 4.5e-11 < B

   1. Initial program 9.1%

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

   3. Simplified 9.1%

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

   4. Taylor expanded in A around -inf 2.1%

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

      1. *-commutative 2.1%

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

   6. Simplified 2.1%

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

   7. Taylor expanded in B around inf 2.5%

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

      1. associate-*r/ 2.5%

         \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]

      2. *-rgt-identity 2.5%

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

   9. Simplified 2.5%

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

4. Final simplification 11.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 4.5 \cdot 10^{-11}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(\left(C \cdot -4\right) \cdot A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{F \cdot A}}{B}\\ \end{array} \]

Alternative 13: 9.3% accurate, 5.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -2 \cdot \frac{\sqrt{F \cdot A}}{B} \end{array} \]

FPCore

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 (* F A)) B)))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	return -2.0 * (sqrt((F * A)) / B);
}

Fortran

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((f * a)) / b)
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	return -2.0 * (Math.sqrt((F * A)) / B);
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	return -2.0 * (math.sqrt((F * A)) / B)

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	return Float64(-2.0 * Float64(sqrt(Float64(F * A)) / B))
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	tmp = -2.0 * (sqrt((F * A)) / B);
end

Wolfram

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[(F * A), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 20.3%

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

3. Simplified 20.3%

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

4. Taylor expanded in A around -inf 12.8%

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

   1. *-commutative 12.8%

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

6. Simplified 12.8%

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

7. Taylor expanded in B around inf 2.3%

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

   1. associate-*r/ 2.3%

      \[\leadsto -2 \cdot \color{blue}{\frac{\sqrt{A \cdot F} \cdot 1}{B}} \]

   2. *-rgt-identity 2.3%

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

9. Simplified 2.3%

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

10. Final simplification 2.3%

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

Reproduce

herbie shell --seed 2023199 
(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))))