ABCF->ab-angle b

Percentage Accurate: 19.0% → 48.7%

Time: 29.6s

Alternatives: 13

Speedup: 5.1×

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: 19.0% 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: 48.7% accurate, 0.3× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\\ t_1 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_2 := {\left(A - C\right)}^{2}\\ t_3 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ t_4 := \frac{-\sqrt{\left(2 \cdot \left(t_3 \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + t_2}\right)}}{t_3}\\ t_5 := \frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B}{C - A}\right)\right)\right)\right)}}{t_0}\\ \mathbf{if}\;t_4 \leq -\infty:\\ \;\;\;\;t_5\\ \mathbf{elif}\;t_4 \leq -5 \cdot 10^{-156}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_1\right) \cdot \left(\left(A + C\right) - \sqrt{t_2 + B \cdot B}\right)\right)}}{t_1}\\ \mathbf{elif}\;t_4 \leq \infty:\\ \;\;\;\;t_5\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \end{array} \]

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 (fma B B (* A (* C -4.0))))
        (t_1 (- (* B B) (* 4.0 (* A C))))
        (t_2 (pow (- A C) 2.0))
        (t_3 (- (pow B 2.0) (* (* 4.0 A) C)))
        (t_4
         (/
          (-
           (sqrt (* (* 2.0 (* t_3 F)) (- (+ A C) (sqrt (+ (pow B 2.0) t_2))))))
          t_3))
        (t_5
         (/
          (-
           (sqrt
            (* 2.0 (* t_0 (* F (+ A (+ A (* -0.5 (/ (* B B) (- C A))))))))))
          t_0)))
   (if (<= t_4 (- INFINITY))
     t_5
     (if (<= t_4 -5e-156)
       (/
        (- (sqrt (* 2.0 (* (* F t_1) (- (+ A C) (sqrt (+ t_2 (* B B))))))))
        t_1)
       (if (<= t_4 INFINITY)
         t_5
         (/ (* (sqrt 2.0) (- (sqrt (* F (- A (hypot A B)))))) B))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = fma(B, B, (A * (C * -4.0)));
	double t_1 = (B * B) - (4.0 * (A * C));
	double t_2 = pow((A - C), 2.0);
	double t_3 = pow(B, 2.0) - ((4.0 * A) * C);
	double t_4 = -sqrt(((2.0 * (t_3 * F)) * ((A + C) - sqrt((pow(B, 2.0) + t_2))))) / t_3;
	double t_5 = -sqrt((2.0 * (t_0 * (F * (A + (A + (-0.5 * ((B * B) / (C - A))))))))) / t_0;
	double tmp;
	if (t_4 <= -((double) INFINITY)) {
		tmp = t_5;
	} else if (t_4 <= -5e-156) {
		tmp = -sqrt((2.0 * ((F * t_1) * ((A + C) - sqrt((t_2 + (B * B))))))) / t_1;
	} else if (t_4 <= ((double) INFINITY)) {
		tmp = t_5;
	} else {
		tmp = (sqrt(2.0) * -sqrt((F * (A - hypot(A, B))))) / B;
	}
	return tmp;
}

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = fma(B, B, Float64(A * Float64(C * -4.0)))
	t_1 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_2 = Float64(A - C) ^ 2.0
	t_3 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	t_4 = Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_3 * F)) * Float64(Float64(A + C) - sqrt(Float64((B ^ 2.0) + t_2)))))) / t_3)
	t_5 = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(A + Float64(A + Float64(-0.5 * Float64(Float64(B * B) / Float64(C - A)))))))))) / t_0)
	tmp = 0.0
	if (t_4 <= Float64(-Inf))
		tmp = t_5;
	elseif (t_4 <= -5e-156)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_1) * Float64(Float64(A + C) - sqrt(Float64(t_2 + Float64(B * B)))))))) / t_1);
	elseif (t_4 <= Inf)
		tmp = t_5;
	else
		tmp = Float64(Float64(sqrt(2.0) * Float64(-sqrt(Float64(F * Float64(A - hypot(A, B)))))) / 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[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$3 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$4 = N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$3 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[B, 2.0], $MachinePrecision] + t$95$2), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$3), $MachinePrecision]}, Block[{t$95$5 = N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(A + N[(A + N[(-0.5 * N[(N[(B * B), $MachinePrecision] / N[(C - A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]}, If[LessEqual[t$95$4, (-Infinity)], t$95$5, If[LessEqual[t$95$4, -5e-156], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$1), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(t$95$2 + N[(B * B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[t$95$4, Infinity], t$95$5, N[(N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]]]]]]]]

Derivation

1. Split input into 3 regimes

2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C))) < -inf.0 or -5.00000000000000007e-156 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C))) < +inf.0

   1. Initial program 14.0%

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

   3. Simplified 26.3%

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

   4. Taylor expanded in B around 0 26.3%

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

      1. unpow2 26.3%

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

   6. Simplified 26.3%

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

   if -inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C))) < -5.00000000000000007e-156

   1. Initial program 98.4%

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

   3. Simplified 98.4%

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

   if +inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 2 (*.f64 (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)) F)) (-.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) 2) (pow.f64 B 2))))))) (-.f64 (pow.f64 B 2) (*.f64 (*.f64 4 A) C)))

   1. Initial program 0.0%

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

   3. Simplified 0.0%

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

   4. Taylor expanded in C around 0 1.5%

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

      1. mul-1-neg 1.5%

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

      2. distribute-rgt-neg-in 1.5%

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

      3. unpow2 1.5%

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

      4. unpow2 1.5%

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

      5. hypot-def 11.8%

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

   6. Simplified 11.8%

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

      1. associate-*l/ 11.8%

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

   8. Applied egg-rr 11.8%

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

4. Final simplification 29.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq -\infty:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B}{C - A}\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq -5 \cdot 10^{-156}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{\left(A - C\right)}^{2} + B \cdot B}\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \leq \infty:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B}{C - A}\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \]

Alternative 2: 48.4% accurate, 1.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)\\ \mathbf{if}\;B \leq 3.4 \cdot 10^{+30}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B}{C - A}\right)\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \end{array} \]

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 (fma B B (* A (* C -4.0)))))
   (if (<= B 3.4e+30)
     (/
      (- (sqrt (* 2.0 (* t_0 (* F (+ A (+ A (* -0.5 (/ (* B B) (- C A))))))))))
      t_0)
     (/ (* (sqrt 2.0) (- (sqrt (* F (- A (hypot A B)))))) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = fma(B, B, (A * (C * -4.0)));
	double tmp;
	if (B <= 3.4e+30) {
		tmp = -sqrt((2.0 * (t_0 * (F * (A + (A + (-0.5 * ((B * B) / (C - A))))))))) / t_0;
	} else {
		tmp = (sqrt(2.0) * -sqrt((F * (A - hypot(A, B))))) / B;
	}
	return tmp;
}

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = fma(B, B, Float64(A * Float64(C * -4.0)))
	tmp = 0.0
	if (B <= 3.4e+30)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(A + Float64(A + Float64(-0.5 * Float64(Float64(B * B) / Float64(C - A)))))))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) * Float64(-sqrt(Float64(F * Float64(A - hypot(A, B)))))) / 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[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.4e+30], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(A + N[(A + N[(-0.5 * N[(N[(B * B), $MachinePrecision] / N[(C - A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 3.4000000000000002e30

   1. Initial program 20.2%

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

   3. Simplified 25.9%

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

   4. Taylor expanded in B around 0 16.7%

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

      1. unpow2 16.7%

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

   6. Simplified 16.7%

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

   if 3.4000000000000002e30 < B

   1. Initial program 13.7%

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

   3. Simplified 13.7%

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

   4. Taylor expanded in C around 0 17.8%

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

      1. mul-1-neg 17.8%

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

      2. distribute-rgt-neg-in 17.8%

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

      3. unpow2 17.8%

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

      4. unpow2 17.8%

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

      5. hypot-def 41.6%

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

   6. Simplified 41.6%

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

      1. associate-*l/ 41.6%

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

   8. Applied egg-rr 41.6%

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

4. Final simplification 21.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.4 \cdot 10^{+30}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right) \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B}{C - A}\right)\right)\right)\right)}}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \]

Alternative 3: 48.0% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ t_2 := A - \mathsf{hypot}\left(A, B\right)\\ \mathbf{if}\;B \leq 2.55 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 1.55 \cdot 10^{+70}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot t_2\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot t_2} \cdot \left(\sqrt{2} \cdot \frac{-1}{B}\right)\\ \end{array} \end{array} \]

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 (* A C))))
        (t_1 (* F t_0))
        (t_2 (- A (hypot A B))))
   (if (<= B 2.55e-46)
     (/ (- (sqrt (* 2.0 (* t_1 (+ A A))))) t_0)
     (if (<= B 1.55e+70)
       (/ (- (sqrt (* 2.0 (* t_1 t_2)))) t_0)
       (* (sqrt (* F t_2)) (* (sqrt 2.0) (/ -1.0 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 * (A * C));
	double t_1 = F * t_0;
	double t_2 = A - hypot(A, B);
	double tmp;
	if (B <= 2.55e-46) {
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 1.55e+70) {
		tmp = -sqrt((2.0 * (t_1 * t_2))) / t_0;
	} else {
		tmp = sqrt((F * t_2)) * (sqrt(2.0) * (-1.0 / 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 * (A * C));
	double t_1 = F * t_0;
	double t_2 = A - Math.hypot(A, B);
	double tmp;
	if (B <= 2.55e-46) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 1.55e+70) {
		tmp = -Math.sqrt((2.0 * (t_1 * t_2))) / t_0;
	} else {
		tmp = Math.sqrt((F * t_2)) * (Math.sqrt(2.0) * (-1.0 / B));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	t_1 = F * t_0
	t_2 = A - math.hypot(A, B)
	tmp = 0
	if B <= 2.55e-46:
		tmp = -math.sqrt((2.0 * (t_1 * (A + A)))) / t_0
	elif B <= 1.55e+70:
		tmp = -math.sqrt((2.0 * (t_1 * t_2))) / t_0
	else:
		tmp = math.sqrt((F * t_2)) * (math.sqrt(2.0) * (-1.0 / 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(A * C)))
	t_1 = Float64(F * t_0)
	t_2 = Float64(A - hypot(A, B))
	tmp = 0.0
	if (B <= 2.55e-46)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + A))))) / t_0);
	elseif (B <= 1.55e+70)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * t_2)))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * t_2)) * Float64(sqrt(2.0) * Float64(-1.0 / 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 * (A * C));
	t_1 = F * t_0;
	t_2 = A - hypot(A, B);
	tmp = 0.0;
	if (B <= 2.55e-46)
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	elseif (B <= 1.55e+70)
		tmp = -sqrt((2.0 * (t_1 * t_2))) / t_0;
	else
		tmp = sqrt((F * t_2)) * (sqrt(2.0) * (-1.0 / 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.55e-46], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 1.55e+70], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * t$95$2), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * t$95$2), $MachinePrecision]], $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] * N[(-1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 2.5499999999999999e-46

   1. Initial program 19.0%

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

   3. Simplified 19.0%

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

   4. Taylor expanded in C around inf 14.1%

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

      1. cancel-sign-sub-inv 14.1%

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

      2. metadata-eval 14.1%

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

      3. *-lft-identity 14.1%

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

   6. Simplified 14.1%

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

   if 2.5499999999999999e-46 < B < 1.55000000000000015e70

   1. Initial program 36.7%

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

   3. Simplified 36.7%

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

   4. Taylor expanded in C around 0 24.5%

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

      1. unpow2 24.5%

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

      2. unpow2 24.5%

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

      3. hypot-def 24.7%

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

   6. Simplified 24.7%

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

   if 1.55000000000000015e70 < B

   1. Initial program 6.0%

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

   3. Simplified 6.0%

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

   4. Taylor expanded in C around 0 15.3%

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

      1. mul-1-neg 15.3%

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

      2. distribute-rgt-neg-in 15.3%

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

      3. unpow2 15.3%

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

      4. unpow2 15.3%

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

      5. hypot-def 44.1%

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

   6. Simplified 44.1%

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

      1. div-inv 44.1%

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

   8. Applied egg-rr 44.1%

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

4. Final simplification 19.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.55 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 1.55 \cdot 10^{+70}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \left(\sqrt{2} \cdot \frac{-1}{B}\right)\\ \end{array} \]

Alternative 4: 48.1% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ t_2 := A - \mathsf{hypot}\left(A, B\right)\\ \mathbf{if}\;B \leq 6.5 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 3 \cdot 10^{+69}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot t_2\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot t_2} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

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 (* A C))))
        (t_1 (* F t_0))
        (t_2 (- A (hypot A B))))
   (if (<= B 6.5e-46)
     (/ (- (sqrt (* 2.0 (* t_1 (+ A A))))) t_0)
     (if (<= B 3e+69)
       (/ (- (sqrt (* 2.0 (* t_1 t_2)))) t_0)
       (* (sqrt (* F t_2)) (/ (- (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 * (A * C));
	double t_1 = F * t_0;
	double t_2 = A - hypot(A, B);
	double tmp;
	if (B <= 6.5e-46) {
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 3e+69) {
		tmp = -sqrt((2.0 * (t_1 * t_2))) / t_0;
	} else {
		tmp = sqrt((F * t_2)) * (-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 * (A * C));
	double t_1 = F * t_0;
	double t_2 = A - Math.hypot(A, B);
	double tmp;
	if (B <= 6.5e-46) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 3e+69) {
		tmp = -Math.sqrt((2.0 * (t_1 * t_2))) / t_0;
	} else {
		tmp = Math.sqrt((F * t_2)) * (-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 * (A * C))
	t_1 = F * t_0
	t_2 = A - math.hypot(A, B)
	tmp = 0
	if B <= 6.5e-46:
		tmp = -math.sqrt((2.0 * (t_1 * (A + A)))) / t_0
	elif B <= 3e+69:
		tmp = -math.sqrt((2.0 * (t_1 * t_2))) / t_0
	else:
		tmp = math.sqrt((F * t_2)) * (-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(A * C)))
	t_1 = Float64(F * t_0)
	t_2 = Float64(A - hypot(A, B))
	tmp = 0.0
	if (B <= 6.5e-46)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + A))))) / t_0);
	elseif (B <= 3e+69)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * t_2)))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * t_2)) * 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 * (A * C));
	t_1 = F * t_0;
	t_2 = A - hypot(A, B);
	tmp = 0.0;
	if (B <= 6.5e-46)
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	elseif (B <= 3e+69)
		tmp = -sqrt((2.0 * (t_1 * t_2))) / t_0;
	else
		tmp = sqrt((F * t_2)) * (-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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 6.5e-46], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 3e+69], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * t$95$2), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * t$95$2), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 6.49999999999999966e-46

   1. Initial program 19.0%

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

   3. Simplified 19.0%

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

   4. Taylor expanded in C around inf 14.1%

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

      1. cancel-sign-sub-inv 14.1%

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

      2. metadata-eval 14.1%

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

      3. *-lft-identity 14.1%

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

   6. Simplified 14.1%

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

   if 6.49999999999999966e-46 < B < 2.99999999999999983e69

   1. Initial program 36.7%

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

   3. Simplified 36.7%

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

   4. Taylor expanded in C around 0 24.5%

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

      1. unpow2 24.5%

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

      2. unpow2 24.5%

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

      3. hypot-def 24.7%

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

   6. Simplified 24.7%

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

   if 2.99999999999999983e69 < B

   1. Initial program 6.0%

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

   3. Simplified 6.0%

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

   4. Taylor expanded in C around 0 15.3%

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

      1. mul-1-neg 15.3%

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

      2. distribute-rgt-neg-in 15.3%

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

      3. unpow2 15.3%

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

      4. unpow2 15.3%

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

      5. hypot-def 44.1%

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

   6. Simplified 44.1%

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

4. Final simplification 19.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6.5 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 3 \cdot 10^{+69}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 5: 48.1% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ t_2 := A - \mathsf{hypot}\left(A, B\right)\\ \mathbf{if}\;B \leq 4 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 1.55 \cdot 10^{+70}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot t_2\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot t_2}\right)}{B}\\ \end{array} \end{array} \]

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 (* A C))))
        (t_1 (* F t_0))
        (t_2 (- A (hypot A B))))
   (if (<= B 4e-46)
     (/ (- (sqrt (* 2.0 (* t_1 (+ A A))))) t_0)
     (if (<= B 1.55e+70)
       (/ (- (sqrt (* 2.0 (* t_1 t_2)))) t_0)
       (/ (* (sqrt 2.0) (- (sqrt (* F t_2)))) 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 * (A * C));
	double t_1 = F * t_0;
	double t_2 = A - hypot(A, B);
	double tmp;
	if (B <= 4e-46) {
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 1.55e+70) {
		tmp = -sqrt((2.0 * (t_1 * t_2))) / t_0;
	} else {
		tmp = (sqrt(2.0) * -sqrt((F * t_2))) / 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 * (A * C));
	double t_1 = F * t_0;
	double t_2 = A - Math.hypot(A, B);
	double tmp;
	if (B <= 4e-46) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 1.55e+70) {
		tmp = -Math.sqrt((2.0 * (t_1 * t_2))) / t_0;
	} else {
		tmp = (Math.sqrt(2.0) * -Math.sqrt((F * t_2))) / B;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	t_1 = F * t_0
	t_2 = A - math.hypot(A, B)
	tmp = 0
	if B <= 4e-46:
		tmp = -math.sqrt((2.0 * (t_1 * (A + A)))) / t_0
	elif B <= 1.55e+70:
		tmp = -math.sqrt((2.0 * (t_1 * t_2))) / t_0
	else:
		tmp = (math.sqrt(2.0) * -math.sqrt((F * t_2))) / 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(A * C)))
	t_1 = Float64(F * t_0)
	t_2 = Float64(A - hypot(A, B))
	tmp = 0.0
	if (B <= 4e-46)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + A))))) / t_0);
	elseif (B <= 1.55e+70)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * t_2)))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) * Float64(-sqrt(Float64(F * t_2)))) / 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 * (A * C));
	t_1 = F * t_0;
	t_2 = A - hypot(A, B);
	tmp = 0.0;
	if (B <= 4e-46)
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	elseif (B <= 1.55e+70)
		tmp = -sqrt((2.0 * (t_1 * t_2))) / t_0;
	else
		tmp = (sqrt(2.0) * -sqrt((F * t_2))) / 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, Block[{t$95$2 = N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 4e-46], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 1.55e+70], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * t$95$2), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(F * t$95$2), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 4.00000000000000009e-46

   1. Initial program 19.0%

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

   3. Simplified 19.0%

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

   4. Taylor expanded in C around inf 14.1%

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

      1. cancel-sign-sub-inv 14.1%

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

      2. metadata-eval 14.1%

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

      3. *-lft-identity 14.1%

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

   6. Simplified 14.1%

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

   if 4.00000000000000009e-46 < B < 1.55000000000000015e70

   1. Initial program 36.7%

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

   3. Simplified 36.7%

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

   4. Taylor expanded in C around 0 24.5%

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

      1. unpow2 24.5%

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

      2. unpow2 24.5%

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

      3. hypot-def 24.7%

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

   6. Simplified 24.7%

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

   if 1.55000000000000015e70 < B

   1. Initial program 6.0%

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

   3. Simplified 6.0%

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

   4. Taylor expanded in C around 0 15.3%

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

      1. mul-1-neg 15.3%

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

      2. distribute-rgt-neg-in 15.3%

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

      3. unpow2 15.3%

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

      4. unpow2 15.3%

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

      5. hypot-def 44.1%

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

   6. Simplified 44.1%

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

      1. associate-*l/ 44.1%

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

   8. Applied egg-rr 44.1%

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

4. Final simplification 19.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 4 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 1.55 \cdot 10^{+70}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \]

Alternative 6: 45.7% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;B \leq 8.6 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 9.8 \cdot 10^{+69}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]

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 (* A C)))) (t_1 (* F t_0)))
   (if (<= B 8.6e-46)
     (/ (- (sqrt (* 2.0 (* t_1 (+ A A))))) t_0)
     (if (<= B 9.8e+69)
       (/ (- (sqrt (* 2.0 (* t_1 (- A (hypot A B)))))) 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 * (A * C));
	double t_1 = F * t_0;
	double tmp;
	if (B <= 8.6e-46) {
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 9.8e+69) {
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / 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 * (A * C));
	double t_1 = F * t_0;
	double tmp;
	if (B <= 8.6e-46) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	} else if (B <= 9.8e+69) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A - Math.hypot(A, B))))) / 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 * (A * C))
	t_1 = F * t_0
	tmp = 0
	if B <= 8.6e-46:
		tmp = -math.sqrt((2.0 * (t_1 * (A + A)))) / t_0
	elif B <= 9.8e+69:
		tmp = -math.sqrt((2.0 * (t_1 * (A - math.hypot(A, B))))) / 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(A * C)))
	t_1 = Float64(F * t_0)
	tmp = 0.0
	if (B <= 8.6e-46)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + A))))) / t_0);
	elseif (B <= 9.8e+69)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A - hypot(A, B)))))) / 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 * (A * C));
	t_1 = F * t_0;
	tmp = 0.0;
	if (B <= 8.6e-46)
		tmp = -sqrt((2.0 * (t_1 * (A + A)))) / t_0;
	elseif (B <= 9.8e+69)
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, If[LessEqual[B, 8.6e-46], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 9.8e+69], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(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 < 8.6000000000000007e-46

   1. Initial program 19.0%

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

   3. Simplified 19.0%

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

   4. Taylor expanded in C around inf 14.1%

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

      1. cancel-sign-sub-inv 14.1%

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

      2. metadata-eval 14.1%

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

      3. *-lft-identity 14.1%

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

   6. Simplified 14.1%

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

   if 8.6000000000000007e-46 < B < 9.7999999999999999e69

   1. Initial program 36.7%

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

   3. Simplified 36.7%

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

   4. Taylor expanded in C around 0 24.5%

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

      1. unpow2 24.5%

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

      2. unpow2 24.5%

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

      3. hypot-def 24.7%

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

   6. Simplified 24.7%

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

   if 9.7999999999999999e69 < B

   1. Initial program 6.0%

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

   3. Simplified 6.0%

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

   4. Taylor expanded in C around 0 15.3%

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

      1. mul-1-neg 15.3%

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

      2. distribute-rgt-neg-in 15.3%

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

      3. unpow2 15.3%

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

      4. unpow2 15.3%

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

      5. hypot-def 44.1%

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

   6. Simplified 44.1%

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

   7. Taylor expanded in A around 0 41.1%

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

      1. mul-1-neg 41.1%

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

      2. unsub-neg 41.1%

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

   9. Simplified 41.1%

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

4. Final simplification 19.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 8.6 \cdot 10^{-46}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 9.8 \cdot 10^{+69}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

Alternative 7: 43.6% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 2.2 \cdot 10^{+52}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]

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 (* A C)))))
   (if (<= B 2.2e+52)
     (/ (- (sqrt (* 2.0 (* (* F t_0) (+ A 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 * (A * C));
	double tmp;
	if (B <= 2.2e+52) {
		tmp = -sqrt((2.0 * ((F * t_0) * (A + 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 * (a * c))
    if (b <= 2.2d+52) then
        tmp = -sqrt((2.0d0 * ((f * t_0) * (a + 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 * (A * C));
	double tmp;
	if (B <= 2.2e+52) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * (A + 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 * (A * C))
	tmp = 0
	if B <= 2.2e+52:
		tmp = -math.sqrt((2.0 * ((F * t_0) * (A + 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(A * C)))
	tmp = 0.0
	if (B <= 2.2e+52)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + 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 * (A * C));
	tmp = 0.0;
	if (B <= 2.2e+52)
		tmp = -sqrt((2.0 * ((F * t_0) * (A + 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.2e+52], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 2.2e52

   1. Initial program 21.3%

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

   3. Simplified 21.3%

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

   4. Taylor expanded in C around inf 14.4%

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

      1. cancel-sign-sub-inv 14.4%

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

      2. metadata-eval 14.4%

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

      3. *-lft-identity 14.4%

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

   6. Simplified 14.4%

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

   if 2.2e52 < B

   1. Initial program 7.7%

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

   3. Simplified 7.7%

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

   4. Taylor expanded in C around 0 16.1%

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

      1. mul-1-neg 16.1%

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

      2. distribute-rgt-neg-in 16.1%

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

      3. unpow2 16.1%

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

      4. unpow2 16.1%

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

      5. hypot-def 41.6%

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

   6. Simplified 41.6%

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

   7. Taylor expanded in A around 0 38.9%

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

      1. mul-1-neg 38.9%

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

      2. unsub-neg 38.9%

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

   9. Simplified 38.9%

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

4. Final simplification 18.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.2 \cdot 10^{+52}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

Alternative 8: 43.1% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 2.8 \cdot 10^{+53}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{B \cdot \left(-F\right)}\right)\\ \end{array} \end{array} \]

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 (* A C)))))
   (if (<= B 2.8e+53)
     (/ (- (sqrt (* 2.0 (* (* F t_0) (+ A A))))) t_0)
     (* (/ (sqrt 2.0) B) (- (sqrt (* B (- F))))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 2.8e+53) {
		tmp = -sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0;
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((B * -F));
	}
	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 * (a * c))
    if (b <= 2.8d+53) then
        tmp = -sqrt((2.0d0 * ((f * t_0) * (a + a)))) / t_0
    else
        tmp = (sqrt(2.0d0) / b) * -sqrt((b * -f))
    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 * (A * C));
	double tmp;
	if (B <= 2.8e+53) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0;
	} else {
		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt((B * -F));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 2.8e+53:
		tmp = -math.sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0
	else:
		tmp = (math.sqrt(2.0) / B) * -math.sqrt((B * -F))
	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(A * C)))
	tmp = 0.0
	if (B <= 2.8e+53)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + A))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(B * Float64(-F)))));
	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 * (A * C));
	tmp = 0.0;
	if (B <= 2.8e+53)
		tmp = -sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0;
	else
		tmp = (sqrt(2.0) / B) * -sqrt((B * -F));
	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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.8e+53], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(B * (-F)), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 2.8e53

   1. Initial program 21.3%

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

   3. Simplified 21.3%

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

   4. Taylor expanded in C around inf 14.4%

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

      1. cancel-sign-sub-inv 14.4%

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

      2. metadata-eval 14.4%

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

      3. *-lft-identity 14.4%

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

   6. Simplified 14.4%

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

   if 2.8e53 < B

   1. Initial program 7.7%

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

   3. Simplified 7.7%

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

   4. Taylor expanded in C around 0 16.1%

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

      1. mul-1-neg 16.1%

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

      2. distribute-rgt-neg-in 16.1%

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

      3. unpow2 16.1%

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

      4. unpow2 16.1%

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

      5. hypot-def 41.6%

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

   6. Simplified 41.6%

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

   7. Taylor expanded in A around 0 40.2%

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

      1. mul-1-neg 40.2%

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

   9. Simplified 40.2%

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

4. Final simplification 18.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.8 \cdot 10^{+53}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{B \cdot \left(-F\right)}\right)\\ \end{array} \]

Alternative 9: 29.6% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 2 \cdot 10^{+90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 (* A C)))))
   (if (<= B 2e+90)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (+ A A)))))) t_0)
     (* -2.0 (/ (sqrt (* A F)) 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 * (A * C));
	double tmp;
	if (B <= 2e+90) {
		tmp = -sqrt((2.0 * (t_0 * (F * (A + A))))) / t_0;
	} else {
		tmp = -2.0 * (sqrt((A * F)) / 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) * (a * c))
    if (b <= 2d+90) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (a + a))))) / t_0
    else
        tmp = (-2.0d0) * (sqrt((a * f)) / 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 * (A * C));
	double tmp;
	if (B <= 2e+90) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (A + A))))) / t_0;
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 2e+90:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (A + A))))) / t_0
	else:
		tmp = -2.0 * (math.sqrt((A * F)) / 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(A * C)))
	tmp = 0.0
	if (B <= 2e+90)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(A + A)))))) / t_0);
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / 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 * (A * C));
	tmp = 0.0;
	if (B <= 2e+90)
		tmp = -sqrt((2.0 * (t_0 * (F * (A + A))))) / t_0;
	else
		tmp = -2.0 * (sqrt((A * F)) / 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2e+90], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 1.99999999999999993e90

   1. Initial program 20.8%

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

   3. Simplified 20.8%

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

   4. Taylor expanded in C around inf 13.8%

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

      1. cancel-sign-sub-inv 13.8%

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

      2. metadata-eval 13.8%

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

      3. *-lft-identity 13.8%

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

   6. Simplified 13.8%

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

      1. distribute-frac-neg 13.8%

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

      2. associate-*l* 14.1%

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

      3. cancel-sign-sub-inv 14.1%

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

      4. metadata-eval 14.1%

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

      5. cancel-sign-sub-inv 14.1%

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

      6. metadata-eval 14.1%

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

   8. Applied egg-rr 14.1%

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

   if 1.99999999999999993e90 < B

   1. Initial program 6.7%

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

   3. Simplified 6.7%

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

   4. Taylor expanded in C around inf 3.6%

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

      1. cancel-sign-sub-inv 3.6%

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

      2. metadata-eval 3.6%

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

      3. *-lft-identity 3.6%

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

   6. Simplified 3.6%

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

   7. Taylor expanded in B around inf 5.2%

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

   8. Taylor expanded in B around inf 6.0%

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

      1. associate-*r/ 6.1%

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

      2. *-commutative 6.1%

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

      3. *-rgt-identity 6.1%

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

   10. Simplified 6.1%

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

4. Final simplification 13.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2 \cdot 10^{+90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 10: 29.6% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 1.15 \cdot 10^{+83}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A + A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 (* A C)))))
   (if (<= B 1.15e+83)
     (/ (- (sqrt (* 2.0 (* (* F t_0) (+ A A))))) t_0)
     (* -2.0 (/ (sqrt (* A F)) 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 * (A * C));
	double tmp;
	if (B <= 1.15e+83) {
		tmp = -sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0;
	} else {
		tmp = -2.0 * (sqrt((A * F)) / 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 * (a * c))
    if (b <= 1.15d+83) then
        tmp = -sqrt((2.0d0 * ((f * t_0) * (a + a)))) / t_0
    else
        tmp = (-2.0d0) * (sqrt((a * f)) / 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 * (A * C));
	double tmp;
	if (B <= 1.15e+83) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0;
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 1.15e+83:
		tmp = -math.sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0
	else:
		tmp = -2.0 * (math.sqrt((A * F)) / 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(A * C)))
	tmp = 0.0
	if (B <= 1.15e+83)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A + A))))) / t_0);
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / 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 * (A * C));
	tmp = 0.0;
	if (B <= 1.15e+83)
		tmp = -sqrt((2.0 * ((F * t_0) * (A + A)))) / t_0;
	else
		tmp = -2.0 * (sqrt((A * F)) / 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.15e+83], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 1.14999999999999997e83

   1. Initial program 21.0%

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

   3. Simplified 21.0%

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

   4. Taylor expanded in C around inf 13.9%

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

      1. cancel-sign-sub-inv 13.9%

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

      2. metadata-eval 13.9%

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

      3. *-lft-identity 13.9%

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

   6. Simplified 13.9%

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

   if 1.14999999999999997e83 < B

   1. Initial program 6.5%

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

   3. Simplified 6.5%

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

   4. Taylor expanded in C around inf 3.5%

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

      1. cancel-sign-sub-inv 3.5%

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

      2. metadata-eval 3.5%

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

      3. *-lft-identity 3.5%

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

   6. Simplified 3.5%

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

   7. Taylor expanded in B around inf 5.1%

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

   8. Taylor expanded in B around inf 5.8%

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

      1. associate-*r/ 5.9%

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

      2. *-commutative 5.9%

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

      3. *-rgt-identity 5.9%

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

   10. Simplified 5.9%

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

4. Final simplification 12.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.15 \cdot 10^{+83}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right)\right) \cdot \left(A + A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 11: 27.1% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 100000000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(-4 \cdot \left(F \cdot \left(A \cdot C\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 100000000000.0)
   (/
    (- (sqrt (* 2.0 (* (+ A A) (* -4.0 (* F (* A C)))))))
    (- (* B B) (* 4.0 (* A C))))
   (* -2.0 (/ (sqrt (* A F)) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 100000000000.0) {
		tmp = -sqrt((2.0 * ((A + A) * (-4.0 * (F * (A * C)))))) / ((B * B) - (4.0 * (A * C)));
	} else {
		tmp = -2.0 * (sqrt((A * F)) / 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 <= 100000000000.0d0) then
        tmp = -sqrt((2.0d0 * ((a + a) * ((-4.0d0) * (f * (a * c)))))) / ((b * b) - (4.0d0 * (a * c)))
    else
        tmp = (-2.0d0) * (sqrt((a * f)) / 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 <= 100000000000.0) {
		tmp = -Math.sqrt((2.0 * ((A + A) * (-4.0 * (F * (A * C)))))) / ((B * B) - (4.0 * (A * C)));
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 100000000000.0:
		tmp = -math.sqrt((2.0 * ((A + A) * (-4.0 * (F * (A * C)))))) / ((B * B) - (4.0 * (A * C)))
	else:
		tmp = -2.0 * (math.sqrt((A * F)) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 100000000000.0)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(A + A) * Float64(-4.0 * Float64(F * Float64(A * C))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 100000000000.0)
		tmp = -sqrt((2.0 * ((A + A) * (-4.0 * (F * (A * C)))))) / ((B * B) - (4.0 * (A * C)));
	else
		tmp = -2.0 * (sqrt((A * F)) / 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, 100000000000.0], N[((-N[Sqrt[N[(2.0 * N[(N[(A + A), $MachinePrecision] * N[(-4.0 * N[(F * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 1e11

   1. Initial program 19.6%

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

   3. Simplified 19.6%

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

   4. Taylor expanded in C around inf 13.9%

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

      1. cancel-sign-sub-inv 13.9%

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

      2. metadata-eval 13.9%

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

      3. *-lft-identity 13.9%

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

   6. Simplified 13.9%

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

   7. Taylor expanded in B around 0 11.9%

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

      1. associate-*r* 13.6%

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

   9. Simplified 13.6%

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

   if 1e11 < B

   1. Initial program 16.7%

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

   3. Simplified 16.7%

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

   4. Taylor expanded in C around inf 6.9%

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

      1. cancel-sign-sub-inv 6.9%

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

      2. metadata-eval 6.9%

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

      3. *-lft-identity 6.9%

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

   6. Simplified 6.9%

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

   7. Taylor expanded in B around inf 6.5%

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

   8. Taylor expanded in B around inf 7.1%

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

      1. associate-*r/ 7.1%

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

      2. *-commutative 7.1%

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

      3. *-rgt-identity 7.1%

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

   10. Simplified 7.1%

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

4. Final simplification 12.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 100000000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(-4 \cdot \left(F \cdot \left(A \cdot C\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 12: 19.4% accurate, 5.1× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 1.15 \cdot 10^{-40}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(-8 \cdot \left(A \cdot A\right)\right) \cdot \left(C \cdot F\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 1.15e-40)
   (/
    (- (sqrt (* 2.0 (* (* -8.0 (* A A)) (* C F)))))
    (- (* B B) (* 4.0 (* A C))))
   (* -2.0 (/ (sqrt (* A F)) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 1.15e-40) {
		tmp = -sqrt((2.0 * ((-8.0 * (A * A)) * (C * F)))) / ((B * B) - (4.0 * (A * C)));
	} else {
		tmp = -2.0 * (sqrt((A * F)) / 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 <= 1.15d-40) then
        tmp = -sqrt((2.0d0 * (((-8.0d0) * (a * a)) * (c * f)))) / ((b * b) - (4.0d0 * (a * c)))
    else
        tmp = (-2.0d0) * (sqrt((a * f)) / 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 <= 1.15e-40) {
		tmp = -Math.sqrt((2.0 * ((-8.0 * (A * A)) * (C * F)))) / ((B * B) - (4.0 * (A * C)));
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 1.15e-40:
		tmp = -math.sqrt((2.0 * ((-8.0 * (A * A)) * (C * F)))) / ((B * B) - (4.0 * (A * C)))
	else:
		tmp = -2.0 * (math.sqrt((A * F)) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 1.15e-40)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(-8.0 * Float64(A * A)) * Float64(C * F))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 1.15e-40)
		tmp = -sqrt((2.0 * ((-8.0 * (A * A)) * (C * F)))) / ((B * B) - (4.0 * (A * C)));
	else
		tmp = -2.0 * (sqrt((A * F)) / 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, 1.15e-40], N[((-N[Sqrt[N[(2.0 * N[(N[(-8.0 * N[(A * A), $MachinePrecision]), $MachinePrecision] * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 1.15e-40

   1. Initial program 19.3%

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

   3. Simplified 19.3%

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

   4. Taylor expanded in C around inf 13.9%

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

      1. cancel-sign-sub-inv 13.9%

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

      2. metadata-eval 13.9%

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

      3. *-lft-identity 13.9%

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

   6. Simplified 13.9%

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

   7. Taylor expanded in B around 0 9.0%

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

      1. associate-*r* 9.0%

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

      2. unpow2 9.0%

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

      3. *-commutative 9.0%

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

   9. Simplified 9.0%

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

   if 1.15e-40 < B

   1. Initial program 18.3%

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

   3. Simplified 18.3%

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

   4. Taylor expanded in C around inf 8.1%

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

      1. cancel-sign-sub-inv 8.1%

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

      2. metadata-eval 8.1%

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

      3. *-lft-identity 8.1%

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

   6. Simplified 8.1%

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

   7. Taylor expanded in B around inf 5.5%

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

   8. Taylor expanded in B around inf 6.0%

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

      1. associate-*r/ 6.0%

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

      2. *-commutative 6.0%

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

      3. *-rgt-identity 6.0%

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

   10. Simplified 6.0%

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

4. Final simplification 8.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.15 \cdot 10^{-40}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(-8 \cdot \left(A \cdot A\right)\right) \cdot \left(C \cdot F\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 13: 8.9% accurate, 5.9× speedup

Math

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

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 (* A F)) B)))

C

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

Python

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

Julia

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

MATLAB

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

Derivation

1. Initial program 19.0%

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

3. Simplified 19.0%

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

4. Taylor expanded in C around inf 12.5%

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

   1. cancel-sign-sub-inv 12.5%

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

   2. metadata-eval 12.5%

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

   3. *-lft-identity 12.5%

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

6. Simplified 12.5%

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

7. Taylor expanded in B around inf 2.6%

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

8. Taylor expanded in B around inf 2.7%

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

   1. associate-*r/ 2.7%

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

   2. *-commutative 2.7%

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

   3. *-rgt-identity 2.7%

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

10. Simplified 2.7%

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

11. Final simplification 2.7%

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

Reproduce

herbie shell --seed 2023282 
(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))))