ABCF->ab-angle b

Percentage Accurate: 18.8% → 48.8%

Time: 22.0s

Alternatives: 12

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: 18.8% 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: 48.8% accurate, 1.2× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 9.5 \cdot 10^{-53}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 5.5 \cdot 10^{+124}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \left(-\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

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 9.5e-53)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (* 2.0 A))))) t_0)
     (if (<= B 5.5e+124)
       (/
        (*
         (sqrt (fma B B (* C (* A -4.0))))
         (- (sqrt (* 2.0 (* F (+ A (- C (hypot (- A C) B))))))))
        (fma B B (* A (* C -4.0))))
       (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 9.5e-53) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	} else if (B <= 5.5e+124) {
		tmp = (sqrt(fma(B, B, (C * (A * -4.0)))) * -sqrt((2.0 * (F * (A + (C - hypot((A - C), B))))))) / fma(B, B, (A * (C * -4.0)));
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (-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)))
	tmp = 0.0
	if (B <= 9.5e-53)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(2.0 * A))))) / t_0);
	elseif (B <= 5.5e+124)
		tmp = Float64(Float64(sqrt(fma(B, B, Float64(C * Float64(A * -4.0)))) * Float64(-sqrt(Float64(2.0 * Float64(F * Float64(A + Float64(C - hypot(Float64(A - C), B)))))))) / fma(B, B, Float64(A * Float64(C * -4.0))));
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

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, 9.5e-53], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 5.5e+124], N[(N[(N[Sqrt[N[(B * B + N[(C * N[(A * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[N[(2.0 * N[(F * N[(A + N[(C - N[Sqrt[N[(A - C), $MachinePrecision] ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / N[(B * B + N[(A * N[(C * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if B < 9.5000000000000008e-53

   1. Initial program 19.1%

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

   3. Simplified 19.1%

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

   4. Taylor expanded in A around -inf 19.7%

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

      1. *-commutative 19.7%

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

   6. Simplified 19.7%

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

   if 9.5000000000000008e-53 < B < 5.49999999999999977e124

   1. Initial program 27.6%

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

   3. Simplified 35.9%

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

      1. sqrt-prod 45.6%

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

      2. associate-*r* 45.6%

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

      3. *-commutative 45.6%

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

      4. associate-*l* 45.6%

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

      5. associate--r- 45.5%

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

      6. +-commutative 45.5%

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

   5. Applied egg-rr 45.5%

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

      1. hypot-def 37.5%

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

      2. unpow2 37.5%

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

      3. unpow2 37.5%

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

      4. +-commutative 37.5%

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

      5. unpow2 37.5%

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

      6. unpow2 37.5%

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

      7. hypot-def 45.5%

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

   7. Simplified 45.5%

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

   if 5.49999999999999977e124 < B

   1. Initial program 0.5%

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

   3. Simplified 0.5%

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

   4. Taylor expanded in C around 0 9.6%

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

      1. mul-1-neg 9.6%

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

      2. *-commutative 9.6%

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

      3. +-commutative 9.6%

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

      4. unpow2 9.6%

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

      5. unpow2 9.6%

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

      6. hypot-def 47.4%

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

   6. Simplified 47.4%

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

4. Final simplification 26.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 9.5 \cdot 10^{-53}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 5.5 \cdot 10^{+124}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right)} \cdot \left(-\sqrt{2 \cdot \left(F \cdot \left(A + \left(C - \mathsf{hypot}\left(A - C, B\right)\right)\right)\right)}\right)}{\mathsf{fma}\left(B, B, A \cdot \left(C \cdot -4\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 2: 47.4% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 6 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

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 6e+20)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (* 2.0 A))))) t_0)
     (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 6e+20) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 6e+20) {
		tmp = -Math.sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(A, B)))) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 6e+20:
		tmp = -math.sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0
	else:
		tmp = math.sqrt((F * (A - math.hypot(A, B)))) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 6e+20)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(2.0 * A))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	tmp = 0.0;
	if (B <= 6e+20)
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	else
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 6e+20], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 6e20

   1. Initial program 19.6%

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

   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 A around -inf 19.1%

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

      1. *-commutative 19.1%

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

   6. Simplified 19.1%

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

   if 6e20 < B

   1. Initial program 7.8%

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

   3. Simplified 7.8%

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

   4. Taylor expanded in C around 0 16.8%

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

      1. mul-1-neg 16.8%

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

      2. *-commutative 16.8%

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

      3. +-commutative 16.8%

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

      4. unpow2 16.8%

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

      5. unpow2 16.8%

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

      6. hypot-def 44.6%

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

   6. Simplified 44.6%

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

4. Final simplification 24.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 3: 35.4% accurate, 2.6× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;C \leq 1.2 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

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 (* t_0 F)))
   (if (<= C 1.2e+42)
     (/ (- (sqrt (* 2.0 (* t_1 (- A (hypot A B)))))) t_0)
     (/
      (-
       (sqrt
        (*
         2.0
         (* t_1 (+ A (fma -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C) A))))))
      t_0))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (C <= 1.2e+42) {
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
	} else {
		tmp = -sqrt((2.0 * (t_1 * (A + fma(-0.5, (((B * B) + ((A * A) - (A * A))) / C), A))))) / t_0;
	}
	return tmp;
}

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_1 = Float64(t_0 * F)
	tmp = 0.0
	if (C <= 1.2e+42)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A - hypot(A, B)))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + fma(-0.5, Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C), A)))))) / t_0);
	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[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[C, 1.2e+42], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision] + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if C < 1.1999999999999999e42

   1. Initial program 21.2%

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

   3. Simplified 21.2%

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

   4. Taylor expanded in C around 0 14.7%

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

      1. +-commutative 14.7%

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

      2. unpow2 14.7%

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

      3. unpow2 14.7%

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

      4. hypot-def 18.2%

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

   6. Simplified 18.2%

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

   if 1.1999999999999999e42 < C

   1. Initial program 2.7%

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

   3. Simplified 2.7%

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

   4. Taylor expanded in C around inf 35.0%

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

      1. associate--l+ 35.0%

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

      2. fma-neg 35.0%

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

      3. associate--l+ 35.0%

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

      4. unpow2 35.0%

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

      5. unpow2 35.0%

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

      6. unpow2 35.0%

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

      7. mul-1-neg 35.0%

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

      8. mul-1-neg 35.0%

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

      9. sqr-neg 35.0%

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

      10. mul-1-neg 35.0%

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

   6. Simplified 35.0%

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

4. Final simplification 22.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 1.2 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \mathsf{fma}\left(-0.5, \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}, A\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

Alternative 4: 29.8% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + \left(A \cdot C\right) \cdot -4\\ t_1 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 5.4 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_1 \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{+117}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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) (* (* A C) -4.0))) (t_1 (- (* B B) (* 4.0 (* A C)))))
   (if (<= B 5.4e+20)
     (/ (- (sqrt (* 2.0 (* (* t_1 F) (* 2.0 A))))) t_1)
     (if (<= B 3.2e+117)
       (/ (- (sqrt (* 2.0 (* t_0 (* F (- C (hypot C B))))))) 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) + ((A * C) * -4.0);
	double t_1 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 5.4e+20) {
		tmp = -sqrt((2.0 * ((t_1 * F) * (2.0 * A)))) / t_1;
	} else if (B <= 3.2e+117) {
		tmp = -sqrt((2.0 * (t_0 * (F * (C - hypot(C, B)))))) / t_0;
	} else {
		tmp = -2.0 * (sqrt((A * F)) / 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) + ((A * C) * -4.0);
	double t_1 = (B * B) - (4.0 * (A * C));
	double tmp;
	if (B <= 5.4e+20) {
		tmp = -Math.sqrt((2.0 * ((t_1 * F) * (2.0 * A)))) / t_1;
	} else if (B <= 3.2e+117) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (C - Math.hypot(C, B)))))) / 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) + ((A * C) * -4.0)
	t_1 = (B * B) - (4.0 * (A * C))
	tmp = 0
	if B <= 5.4e+20:
		tmp = -math.sqrt((2.0 * ((t_1 * F) * (2.0 * A)))) / t_1
	elif B <= 3.2e+117:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (C - math.hypot(C, B)))))) / 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(Float64(A * C) * -4.0))
	t_1 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 5.4e+20)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_1 * F) * Float64(2.0 * A))))) / t_1);
	elseif (B <= 3.2e+117)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(C - hypot(C, B))))))) / 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) + ((A * C) * -4.0);
	t_1 = (B * B) - (4.0 * (A * C));
	tmp = 0.0;
	if (B <= 5.4e+20)
		tmp = -sqrt((2.0 * ((t_1 * F) * (2.0 * A)))) / t_1;
	elseif (B <= 3.2e+117)
		tmp = -sqrt((2.0 * (t_0 * (F * (C - hypot(C, B)))))) / 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[(N[(A * C), $MachinePrecision] * -4.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 5.4e+20], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$1 * F), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 3.2e+117], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(C - N[Sqrt[C ^ 2 + B ^ 2], $MachinePrecision]), $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 3 regimes

2. if B < 5.4e20

   1. Initial program 19.6%

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

   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 A around -inf 19.1%

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

      1. *-commutative 19.1%

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

   6. Simplified 19.1%

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

   if 5.4e20 < B < 3.20000000000000005e117

   1. Initial program 31.8%

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

   3. Simplified 31.8%

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

   4. Taylor expanded in A around 0 32.1%

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

      1. +-commutative 32.1%

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

      2. unpow2 32.1%

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

      3. unpow2 32.1%

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

      4. hypot-def 32.4%

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

   6. Simplified 32.4%

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

      1. distribute-frac-neg 32.4%

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

      2. associate-*l* 32.6%

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

      3. cancel-sign-sub-inv 32.6%

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

      4. metadata-eval 32.6%

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

      5. cancel-sign-sub-inv 32.6%

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

      6. metadata-eval 32.6%

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

   8. Applied egg-rr 32.6%

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

   if 3.20000000000000005e117 < B

   1. Initial program 0.6%

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

   3. Simplified 0.6%

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

   4. Taylor expanded in A around -inf 0.3%

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

      1. *-commutative 0.3%

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

   6. Simplified 0.3%

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

   7. Taylor expanded in B around inf 4.1%

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

      1. associate-*r/ 4.1%

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

      2. *-rgt-identity 4.1%

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

      3. *-commutative 4.1%

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

   9. Simplified 4.1%

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

4. Final simplification 17.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 5.4 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 3.2 \cdot 10^{+117}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + \left(A \cdot C\right) \cdot -4\right) \cdot \left(F \cdot \left(C - \mathsf{hypot}\left(C, B\right)\right)\right)\right)}}{B \cdot B + \left(A \cdot C\right) \cdot -4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 5: 32.5% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;C \leq 1.12 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \end{array} \end{array} \]

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 (* t_0 F)))
   (if (<= C 1.12e+42)
     (/ (- (sqrt (* 2.0 (* t_1 (- A (hypot A B)))))) t_0)
     (/ (- (sqrt (* 2.0 (* t_1 (* 2.0 A))))) t_0))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (C <= 1.12e+42) {
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
	} else {
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	}
	return tmp;
}

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 = t_0 * F;
	double tmp;
	if (C <= 1.12e+42) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A - Math.hypot(A, B))))) / t_0;
	} else {
		tmp = -Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	}
	return tmp;
}

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 = t_0 * F
	tmp = 0
	if C <= 1.12e+42:
		tmp = -math.sqrt((2.0 * (t_1 * (A - math.hypot(A, B))))) / t_0
	else:
		tmp = -math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0
	return tmp

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(t_0 * F)
	tmp = 0.0
	if (C <= 1.12e+42)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A - hypot(A, B)))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A))))) / t_0);
	end
	return tmp
end

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 = t_0 * F;
	tmp = 0.0;
	if (C <= 1.12e+42)
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
	else
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	end
	tmp_2 = tmp;
end

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[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[C, 1.12e+42], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if C < 1.12e42

   1. Initial program 21.2%

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

   3. Simplified 21.2%

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

   4. Taylor expanded in C around 0 14.7%

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

      1. +-commutative 14.7%

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

      2. unpow2 14.7%

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

      3. unpow2 14.7%

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

      4. hypot-def 18.2%

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

   6. Simplified 18.2%

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

   if 1.12e42 < C

   1. Initial program 2.7%

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

   3. Simplified 2.7%

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

   4. Taylor expanded in A around -inf 30.8%

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

      1. *-commutative 30.8%

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

   6. Simplified 30.8%

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

4. Final simplification 21.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 1.12 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

Alternative 6: 29.8% accurate, 2.8× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 7 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 1.8 \cdot 10^{+117}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(C - \mathsf{hypot}\left(C, B\right)\right) \cdot \left(\left(B \cdot B\right) \cdot F\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 7e+20)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (* 2.0 A))))) t_0)
     (if (<= B 1.8e+117)
       (/ (- (sqrt (* 2.0 (* (- C (hypot C B)) (* (* B B) F))))) 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 <= 7e+20) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	} else if (B <= 1.8e+117) {
		tmp = -sqrt((2.0 * ((C - hypot(C, B)) * ((B * B) * F)))) / t_0;
	} else {
		tmp = -2.0 * (sqrt((A * F)) / 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 tmp;
	if (B <= 7e+20) {
		tmp = -Math.sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	} else if (B <= 1.8e+117) {
		tmp = -Math.sqrt((2.0 * ((C - Math.hypot(C, B)) * ((B * B) * F)))) / 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 <= 7e+20:
		tmp = -math.sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0
	elif B <= 1.8e+117:
		tmp = -math.sqrt((2.0 * ((C - math.hypot(C, B)) * ((B * B) * F)))) / 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 <= 7e+20)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(2.0 * A))))) / t_0);
	elseif (B <= 1.8e+117)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(C - hypot(C, B)) * Float64(Float64(B * B) * F))))) / 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 <= 7e+20)
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * A)))) / t_0;
	elseif (B <= 1.8e+117)
		tmp = -sqrt((2.0 * ((C - hypot(C, B)) * ((B * B) * F)))) / 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, 7e+20], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 1.8e+117], N[((-N[Sqrt[N[(2.0 * N[(N[(C - N[Sqrt[C ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision] * N[(N[(B * B), $MachinePrecision] * F), $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 3 regimes

2. if B < 7e20

   1. Initial program 19.6%

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

   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 A around -inf 19.1%

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

      1. *-commutative 19.1%

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

   6. Simplified 19.1%

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

   if 7e20 < B < 1.80000000000000006e117

   1. Initial program 31.8%

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

   3. Simplified 31.8%

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

   4. Taylor expanded in A around 0 32.1%

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

      1. +-commutative 32.1%

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

      2. unpow2 32.1%

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

      3. unpow2 32.1%

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

      4. hypot-def 32.4%

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

   6. Simplified 32.4%

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

   7. Taylor expanded in B around inf 32.1%

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

      1. unpow2 32.1%

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

   9. Simplified 32.1%

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

   if 1.80000000000000006e117 < B

   1. Initial program 0.6%

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

   3. Simplified 0.6%

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

   4. Taylor expanded in A around -inf 0.3%

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

      1. *-commutative 0.3%

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

   6. Simplified 0.3%

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

   7. Taylor expanded in B around inf 4.1%

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

      1. associate-*r/ 4.1%

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

      2. *-rgt-identity 4.1%

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

      3. *-commutative 4.1%

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

   9. Simplified 4.1%

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

4. Final simplification 17.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 7 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 1.8 \cdot 10^{+117}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(C - \mathsf{hypot}\left(C, B\right)\right) \cdot \left(\left(B \cdot B\right) \cdot F\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 7: 29.8% accurate, 4.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;B \leq 8.5 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{elif}\;B \leq 1.5 \cdot 10^{+117}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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)))) (t_1 (* t_0 F)))
   (if (<= B 8.5e+20)
     (/ (- (sqrt (* 2.0 (* t_1 (* 2.0 A))))) t_0)
     (if (<= B 1.5e+117)
       (/ (- (sqrt (* 2.0 (* t_1 (- (+ A C) B))))) 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 t_1 = t_0 * F;
	double tmp;
	if (B <= 8.5e+20) {
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	} else if (B <= 1.5e+117) {
		tmp = -sqrt((2.0 * (t_1 * ((A + C) - B)))) / 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) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (a * c))
    t_1 = t_0 * f
    if (b <= 8.5d+20) then
        tmp = -sqrt((2.0d0 * (t_1 * (2.0d0 * a)))) / t_0
    else if (b <= 1.5d+117) then
        tmp = -sqrt((2.0d0 * (t_1 * ((a + c) - b)))) / 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 t_1 = t_0 * F;
	double tmp;
	if (B <= 8.5e+20) {
		tmp = -Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	} else if (B <= 1.5e+117) {
		tmp = -Math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / 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))
	t_1 = t_0 * F
	tmp = 0
	if B <= 8.5e+20:
		tmp = -math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0
	elif B <= 1.5e+117:
		tmp = -math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / 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)))
	t_1 = Float64(t_0 * F)
	tmp = 0.0
	if (B <= 8.5e+20)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A))))) / t_0);
	elseif (B <= 1.5e+117)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(Float64(A + C) - B))))) / 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));
	t_1 = t_0 * F;
	tmp = 0.0;
	if (B <= 8.5e+20)
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	elseif (B <= 1.5e+117)
		tmp = -sqrt((2.0 * (t_1 * ((A + C) - B)))) / 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]}, Block[{t$95$1 = N[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[B, 8.5e+20], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[B, 1.5e+117], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(N[(A + C), $MachinePrecision] - B), $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 3 regimes

2. if B < 8.5e20

   1. Initial program 19.6%

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

   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 A around -inf 19.1%

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

      1. *-commutative 19.1%

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

   6. Simplified 19.1%

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

   if 8.5e20 < B < 1.5e117

   1. Initial program 31.8%

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

   3. Simplified 31.8%

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

   4. Taylor expanded in B around inf 24.9%

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

   if 1.5e117 < B

   1. Initial program 0.6%

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

   3. Simplified 0.6%

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

   4. Taylor expanded in A around -inf 0.3%

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

      1. *-commutative 0.3%

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

   6. Simplified 0.3%

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

   7. Taylor expanded in B around inf 4.1%

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

      1. associate-*r/ 4.1%

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

      2. *-rgt-identity 4.1%

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

      3. *-commutative 4.1%

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

   9. Simplified 4.1%

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

4. Final simplification 16.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 8.5 \cdot 10^{+20}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 1.5 \cdot 10^{+117}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 8: 29.0% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + \left(A \cdot C\right) \cdot -4\\ \mathbf{if}\;B \leq 2.05 \cdot 10^{+62}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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) (* (* A C) -4.0))))
   (if (<= B 2.05e+62)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 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) + ((A * C) * -4.0);
	double tmp;
	if (B <= 2.05e+62) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * 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) + ((a * c) * (-4.0d0))
    if (b <= 2.05d+62) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (2.0d0 * 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) + ((A * C) * -4.0);
	double tmp;
	if (B <= 2.05e+62) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * 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) + ((A * C) * -4.0)
	tmp = 0
	if B <= 2.05e+62:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * 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(Float64(A * C) * -4.0))
	tmp = 0.0
	if (B <= 2.05e+62)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * 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) + ((A * C) * -4.0);
	tmp = 0.0;
	if (B <= 2.05e+62)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * 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[(N[(A * C), $MachinePrecision] * -4.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.05e+62], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 2.04999999999999992e62

   1. Initial program 19.7%

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

   3. Simplified 19.7%

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

   4. Taylor expanded in A around -inf 19.3%

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

      1. *-commutative 19.3%

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

   6. Simplified 19.3%

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

      1. distribute-frac-neg 19.3%

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

      2. associate-*l* 19.0%

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

      3. cancel-sign-sub-inv 19.0%

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

      4. metadata-eval 19.0%

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

      5. *-commutative 19.0%

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

      6. cancel-sign-sub-inv 19.0%

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

      7. metadata-eval 19.0%

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

   8. Applied egg-rr 19.0%

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

   if 2.04999999999999992e62 < B

   1. Initial program 6.4%

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

   3. Simplified 6.4%

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

   4. Taylor expanded in A around -inf 0.7%

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

      1. *-commutative 0.7%

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

   6. Simplified 0.7%

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

   7. Taylor expanded in B around inf 3.9%

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

      1. associate-*r/ 3.9%

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

      2. *-rgt-identity 3.9%

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

      3. *-commutative 3.9%

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

   9. Simplified 3.9%

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

4. Final simplification 15.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.05 \cdot 10^{+62}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + \left(A \cdot C\right) \cdot -4\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + \left(A \cdot C\right) \cdot -4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 9: 28.8% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 6.5 \cdot 10^{+61}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 6.5e+61)
     (/ (- (sqrt (* 2.0 (* (* t_0 F) (* 2.0 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 <= 6.5e+61) {
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * 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 <= 6.5d+61) then
        tmp = -sqrt((2.0d0 * ((t_0 * f) * (2.0d0 * 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 <= 6.5e+61) {
		tmp = -Math.sqrt((2.0 * ((t_0 * F) * (2.0 * 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 <= 6.5e+61:
		tmp = -math.sqrt((2.0 * ((t_0 * F) * (2.0 * 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 <= 6.5e+61)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(2.0 * 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 <= 6.5e+61)
		tmp = -sqrt((2.0 * ((t_0 * F) * (2.0 * 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, 6.5e+61], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 6.4999999999999996e61

   1. Initial program 19.7%

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

   3. Simplified 19.7%

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

   4. Taylor expanded in A around -inf 19.3%

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

      1. *-commutative 19.3%

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

   6. Simplified 19.3%

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

   if 6.4999999999999996e61 < B

   1. Initial program 6.4%

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

   3. Simplified 6.4%

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

   4. Taylor expanded in A around -inf 0.7%

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

      1. *-commutative 0.7%

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

   6. Simplified 0.7%

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

   7. Taylor expanded in B around inf 3.9%

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

      1. associate-*r/ 3.9%

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

      2. *-rgt-identity 3.9%

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

      3. *-commutative 3.9%

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

   9. Simplified 3.9%

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

4. Final simplification 16.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6.5 \cdot 10^{+61}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 10: 19.8% accurate, 5.1× speedup

Math

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

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

C

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

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 = -sqrt((2.0d0 * ((2.0d0 * a) * ((a * (-4.0d0)) * (c * f))))) / ((b * b) - (4.0d0 * (a * c)))
end function

Java

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

Python

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

Julia

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

MATLAB

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

Derivation

1. Initial program 17.0%

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

3. Simplified 17.0%

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

4. Taylor expanded in A around -inf 15.5%

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

   1. *-commutative 15.5%

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

6. Simplified 15.5%

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

7. Taylor expanded in B around 0 11.8%

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

   1. associate-*r* 11.8%

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

   2. *-commutative 11.8%

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

9. Simplified 11.8%

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

10. Final simplification 11.8%

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

Alternative 11: 15.6% accurate, 5.2× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \frac{-\sqrt{2 \cdot \left(\left(C \cdot F\right) \cdot \left(\left(A \cdot A\right) \cdot -8\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)} \end{array} \]

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

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	return -sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (A * C)));
}

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 = -sqrt((2.0d0 * ((c * f) * ((a * a) * (-8.0d0))))) / ((b * b) - (4.0d0 * (a * c)))
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	return -Math.sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (A * C)));
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	return -math.sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (A * C)))

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	return Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(C * F) * Float64(Float64(A * A) * -8.0))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(A * C))))
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp = code(A, B, C, F)
	tmp = -sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (A * C)));
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[((-N[Sqrt[N[(2.0 * N[(N[(C * F), $MachinePrecision] * N[(N[(A * A), $MachinePrecision] * -8.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 17.0%

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

3. Simplified 17.0%

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

4. Taylor expanded in A around -inf 15.5%

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

   1. *-commutative 15.5%

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

6. Simplified 15.5%

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

7. Taylor expanded in B around 0 9.6%

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

   1. associate-*r* 9.6%

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

   2. unpow2 9.6%

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

   3. *-commutative 9.6%

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

9. Simplified 9.6%

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

10. Final simplification 9.6%

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

Alternative 12: 9.4% 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 17.0%

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

3. Simplified 17.0%

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

4. Taylor expanded in A around -inf 15.5%

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

   1. *-commutative 15.5%

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

6. Simplified 15.5%

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

7. Taylor expanded in B around inf 3.8%

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

   1. associate-*r/ 3.8%

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

   2. *-rgt-identity 3.8%

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

   3. *-commutative 3.8%

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

9. Simplified 3.8%

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

10. Final simplification 3.8%

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

Reproduce

herbie shell --seed 2023207 
(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))))