ABCF->ab-angle b

Percentage Accurate: 18.5% → 46.0%

Time: 24.6s

Alternatives: 15

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: 18.5% 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: 46.0% accurate, 1.2× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := A - \mathsf{hypot}\left(A, B\right)\\ t_1 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-285}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)}\right)}{t_1}\\ \mathbf{elif}\;{B}^{2} \leq 10^{+30}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_1\right) \cdot t_0\right)}}{t_1}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot t_0} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- A (hypot A B))) (t_1 (- (* B B) (* 4.0 (* C A)))))
   (if (<= (pow B 2.0) 2e-285)
     (/
      (* (sqrt 2.0) (- (sqrt (* (fma B B (* C (* A -4.0))) (* F (* 2.0 A))))))
      t_1)
     (if (<= (pow B 2.0) 1e+30)
       (/ (- (sqrt (* 2.0 (* (* F t_1) t_0)))) t_1)
       (/ (* (sqrt (* F t_0)) (- (sqrt 2.0))) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = A - hypot(A, B);
	double t_1 = (B * B) - (4.0 * (C * A));
	double tmp;
	if (pow(B, 2.0) <= 2e-285) {
		tmp = (sqrt(2.0) * -sqrt((fma(B, B, (C * (A * -4.0))) * (F * (2.0 * A))))) / t_1;
	} else if (pow(B, 2.0) <= 1e+30) {
		tmp = -sqrt((2.0 * ((F * t_1) * t_0))) / t_1;
	} else {
		tmp = (sqrt((F * t_0)) * -sqrt(2.0)) / B;
	}
	return tmp;
}

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(A - hypot(A, B))
	t_1 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	tmp = 0.0
	if ((B ^ 2.0) <= 2e-285)
		tmp = Float64(Float64(sqrt(2.0) * Float64(-sqrt(Float64(fma(B, B, Float64(C * Float64(A * -4.0))) * Float64(F * Float64(2.0 * A)))))) / t_1);
	elseif ((B ^ 2.0) <= 1e+30)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_1) * t_0)))) / t_1);
	else
		tmp = Float64(Float64(sqrt(Float64(F * t_0)) * 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[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B, 2.0], $MachinePrecision], 2e-285], N[(N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(N[(B * B + N[(C * N[(A * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / t$95$1), $MachinePrecision], If[LessEqual[N[Power[B, 2.0], $MachinePrecision], 1e+30], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$1), $MachinePrecision] * t$95$0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], N[(N[(N[Sqrt[N[(F * t$95$0), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if (pow.f64 B 2) < 2.00000000000000015e-285

   1. Initial program 16.4%

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

   3. Simplified 16.4%

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

   4. Taylor expanded in A around -inf 20.3%

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

      1. *-commutative 20.3%

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

   6. Simplified 20.3%

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

      1. sqrt-prod 20.2%

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

      2. associate-*l* 24.8%

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

      3. cancel-sign-sub-inv 24.8%

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

      4. metadata-eval 24.8%

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

      5. associate-*r* 24.8%

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

      6. *-commutative 24.8%

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

      7. *-commutative 24.8%

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

      8. fma-udef 24.8%

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

   8. Applied egg-rr 24.8%

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

   if 2.00000000000000015e-285 < (pow.f64 B 2) < 1e30

   1. Initial program 42.9%

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

   3. Simplified 42.9%

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

   4. Taylor expanded in C around 0 35.0%

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

      1. +-commutative 35.0%

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

      2. unpow2 35.0%

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

      3. unpow2 35.0%

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

      4. hypot-def 39.1%

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

   6. Simplified 39.1%

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

   if 1e30 < (pow.f64 B 2)

   1. Initial program 10.9%

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

   3. Simplified 10.9%

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

   4. Taylor expanded in C around 0 7.1%

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

      1. mul-1-neg 7.1%

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

      2. *-commutative 7.1%

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

      3. +-commutative 7.1%

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

      4. unpow2 7.1%

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

      5. unpow2 7.1%

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

      6. hypot-def 22.8%

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

   6. Simplified 22.8%

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

      1. associate-*l/ 22.8%

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

   8. Applied egg-rr 22.8%

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

4. Final simplification 28.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-285}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{\mathsf{fma}\left(B, B, C \cdot \left(A \cdot -4\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)}\right)}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;{B}^{2} \leq 10^{+30}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \]

Alternative 2: 45.8% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := A - \mathsf{hypot}\left(A, B\right)\\ t_2 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 1.15 \cdot 10^{-142}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_2 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_2}\\ \mathbf{elif}\;B \leq 1.9 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot t_1\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot t_1} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A))))
        (t_1 (- A (hypot A B)))
        (t_2 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= B 1.15e-142)
     (/ (- (sqrt (* 2.0 (* t_2 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_2)
     (if (<= B 1.9e+16)
       (/ (- (sqrt (* 2.0 (* (* F t_0) t_1)))) t_0)
       (* (sqrt (* F t_1)) (/ (- (sqrt 2.0)) B))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = A - hypot(A, B);
	double t_2 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 1.15e-142) {
		tmp = -sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2;
	} else if (B <= 1.9e+16) {
		tmp = -sqrt((2.0 * ((F * t_0) * t_1))) / t_0;
	} else {
		tmp = sqrt((F * t_1)) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = A - Math.hypot(A, B);
	double t_2 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 1.15e-142) {
		tmp = -Math.sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2;
	} else if (B <= 1.9e+16) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * t_1))) / t_0;
	} else {
		tmp = Math.sqrt((F * t_1)) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = A - math.hypot(A, B)
	t_2 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if B <= 1.15e-142:
		tmp = -math.sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2
	elif B <= 1.9e+16:
		tmp = -math.sqrt((2.0 * ((F * t_0) * t_1))) / t_0
	else:
		tmp = math.sqrt((F * t_1)) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(A - hypot(A, B))
	t_2 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 1.15e-142)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_2 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_2);
	elseif (B <= 1.9e+16)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * t_1)))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * t_1)) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = A - hypot(A, B);
	t_2 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (B <= 1.15e-142)
		tmp = -sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2;
	elseif (B <= 1.9e+16)
		tmp = -sqrt((2.0 * ((F * t_0) * t_1))) / t_0;
	else
		tmp = sqrt((F * t_1)) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.15e-142], N[((-N[Sqrt[N[(2.0 * N[(t$95$2 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$2), $MachinePrecision], If[LessEqual[B, 1.9e+16], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * t$95$1), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 1.15000000000000001e-142

   1. Initial program 22.4%

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

   3. Simplified 22.4%

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

      1. distribute-frac-neg 22.4%

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

   5. Applied egg-rr 31.7%

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

   6. Taylor expanded in A around -inf 16.8%

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

      1. unpow2 16.8%

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

   8. Simplified 16.8%

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

   if 1.15000000000000001e-142 < B < 1.9e16

   1. Initial program 39.6%

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

   3. Simplified 39.6%

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

   4. Taylor expanded in C around 0 31.2%

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

      1. +-commutative 31.2%

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

      2. unpow2 31.2%

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

      3. unpow2 31.2%

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

      4. hypot-def 31.6%

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

   6. Simplified 31.6%

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

   if 1.9e16 < B

   1. Initial program 7.9%

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

   3. Simplified 7.9%

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

   4. Taylor expanded in C around 0 13.2%

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

      1. mul-1-neg 13.2%

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

      2. *-commutative 13.2%

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

      3. +-commutative 13.2%

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

      4. unpow2 13.2%

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

      5. unpow2 13.2%

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

      6. hypot-def 43.4%

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

   6. Simplified 43.4%

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

4. Final simplification 25.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.15 \cdot 10^{-142}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 1.9 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 3: 45.8% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := A - \mathsf{hypot}\left(A, B\right)\\ t_2 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 1.15 \cdot 10^{-142}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_2 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_2}\\ \mathbf{elif}\;B \leq 1.5 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot t_1\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot t_1} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A))))
        (t_1 (- A (hypot A B)))
        (t_2 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= B 1.15e-142)
     (/ (- (sqrt (* 2.0 (* t_2 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_2)
     (if (<= B 1.5e+16)
       (/ (- (sqrt (* 2.0 (* (* F t_0) t_1)))) t_0)
       (/ (* (sqrt (* F t_1)) (- (sqrt 2.0))) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = A - hypot(A, B);
	double t_2 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 1.15e-142) {
		tmp = -sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2;
	} else if (B <= 1.5e+16) {
		tmp = -sqrt((2.0 * ((F * t_0) * t_1))) / t_0;
	} else {
		tmp = (sqrt((F * t_1)) * -sqrt(2.0)) / B;
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = A - Math.hypot(A, B);
	double t_2 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 1.15e-142) {
		tmp = -Math.sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2;
	} else if (B <= 1.5e+16) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * t_1))) / t_0;
	} else {
		tmp = (Math.sqrt((F * t_1)) * -Math.sqrt(2.0)) / B;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = A - math.hypot(A, B)
	t_2 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if B <= 1.15e-142:
		tmp = -math.sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2
	elif B <= 1.5e+16:
		tmp = -math.sqrt((2.0 * ((F * t_0) * t_1))) / t_0
	else:
		tmp = (math.sqrt((F * t_1)) * -math.sqrt(2.0)) / B
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(A - hypot(A, B))
	t_2 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 1.15e-142)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_2 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_2);
	elseif (B <= 1.5e+16)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * t_1)))) / t_0);
	else
		tmp = Float64(Float64(sqrt(Float64(F * t_1)) * Float64(-sqrt(2.0))) / B);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = A - hypot(A, B);
	t_2 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (B <= 1.15e-142)
		tmp = -sqrt((2.0 * (t_2 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_2;
	elseif (B <= 1.5e+16)
		tmp = -sqrt((2.0 * ((F * t_0) * t_1))) / t_0;
	else
		tmp = (sqrt((F * t_1)) * -sqrt(2.0)) / B;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.15e-142], N[((-N[Sqrt[N[(2.0 * N[(t$95$2 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$2), $MachinePrecision], If[LessEqual[B, 1.5e+16], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[N[(F * t$95$1), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 1.15000000000000001e-142

   1. Initial program 22.4%

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

   3. Simplified 22.4%

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

      1. distribute-frac-neg 22.4%

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

   5. Applied egg-rr 31.7%

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

   6. Taylor expanded in A around -inf 16.8%

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

      1. unpow2 16.8%

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

   8. Simplified 16.8%

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

   if 1.15000000000000001e-142 < B < 1.5e16

   1. Initial program 39.6%

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

   3. Simplified 39.6%

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

   4. Taylor expanded in C around 0 31.2%

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

      1. +-commutative 31.2%

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

      2. unpow2 31.2%

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

      3. unpow2 31.2%

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

      4. hypot-def 31.6%

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

   6. Simplified 31.6%

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

   if 1.5e16 < B

   1. Initial program 7.9%

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

   3. Simplified 7.9%

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

   4. Taylor expanded in C around 0 13.2%

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

      1. mul-1-neg 13.2%

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

      2. *-commutative 13.2%

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

      3. +-commutative 13.2%

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

      4. unpow2 13.2%

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

      5. unpow2 13.2%

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

      6. hypot-def 43.4%

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

   6. Simplified 43.4%

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

      1. associate-*l/ 43.6%

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

   8. Applied egg-rr 43.6%

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

4. Final simplification 25.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.15 \cdot 10^{-142}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 1.5 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \]

Alternative 4: 43.0% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 7.2 \cdot 10^{-143}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 1.3 \cdot 10^{+32}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= B 7.2e-143)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_1)
     (if (<= B 1.3e+32)
       (/ (- (sqrt (* 2.0 (* (* F t_0) (- A (hypot A B)))))) t_0)
       (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A B)))))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 7.2e-143) {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 1.3e+32) {
		tmp = -sqrt((2.0 * ((F * t_0) * (A - hypot(A, B))))) / t_0;
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 7.2e-143) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 1.3e+32) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * (A - Math.hypot(A, B))))) / t_0;
	} else {
		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt((F * (A - B)));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if B <= 7.2e-143:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1
	elif B <= 1.3e+32:
		tmp = -math.sqrt((2.0 * ((F * t_0) * (A - math.hypot(A, B))))) / t_0
	else:
		tmp = (math.sqrt(2.0) / B) * -math.sqrt((F * (A - B)))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 7.2e-143)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_1);
	elseif (B <= 1.3e+32)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(A - hypot(A, B)))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(F * Float64(A - B)))));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (B <= 7.2e-143)
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	elseif (B <= 1.3e+32)
		tmp = -sqrt((2.0 * ((F * t_0) * (A - hypot(A, B))))) / t_0;
	else
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 7.2e-143], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 1.3e+32], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 7.1999999999999996e-143

   1. Initial program 22.4%

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

   3. Simplified 22.4%

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

      1. distribute-frac-neg 22.4%

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

   5. Applied egg-rr 31.7%

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

   6. Taylor expanded in A around -inf 16.8%

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

      1. unpow2 16.8%

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

   8. Simplified 16.8%

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

   if 7.1999999999999996e-143 < B < 1.3000000000000001e32

   1. Initial program 37.5%

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

   3. Simplified 37.5%

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

   4. Taylor expanded in C around 0 29.5%

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

      1. +-commutative 29.5%

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

      2. unpow2 29.5%

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

      3. unpow2 29.5%

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

      4. hypot-def 29.9%

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

   6. Simplified 29.9%

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

   if 1.3000000000000001e32 < B

   1. Initial program 8.2%

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

   3. Simplified 8.2%

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

   4. Taylor expanded in C around 0 13.5%

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

      1. mul-1-neg 13.5%

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

      2. *-commutative 13.5%

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

      3. +-commutative 13.5%

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

      4. unpow2 13.5%

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

      5. unpow2 13.5%

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

      6. hypot-def 44.8%

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

   6. Simplified 44.8%

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

   7. Taylor expanded in A around 0 41.2%

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

      1. mul-1-neg 41.2%

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

      2. unsub-neg 41.2%

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

   9. Simplified 41.2%

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

4. Final simplification 24.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 7.2 \cdot 10^{-143}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 1.3 \cdot 10^{+32}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

Alternative 5: 42.8% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 6 \cdot 10^{-107}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 1.75 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot t_0\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= B 6e-107)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_1)
     (if (<= B 1.75e+16)
       (/ (- (sqrt (* 2.0 (* (* 2.0 A) (* F t_0))))) t_0)
       (* (/ (sqrt 2.0) B) (- (sqrt (* F (- A B)))))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 6e-107) {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 1.75e+16) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	} else {
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (c * a))
    t_1 = (b * b) + ((-4.0d0) * (c * a))
    if (b <= 6d-107) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (a + (a + (0.5d0 * ((b * b) / a)))))))) / t_1
    else if (b <= 1.75d+16) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * (f * t_0)))) / t_0
    else
        tmp = (sqrt(2.0d0) / b) * -sqrt((f * (a - b)))
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 6e-107) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 1.75e+16) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	} else {
		tmp = (Math.sqrt(2.0) / B) * -Math.sqrt((F * (A - B)));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if B <= 6e-107:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1
	elif B <= 1.75e+16:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0
	else:
		tmp = (math.sqrt(2.0) / B) * -math.sqrt((F * (A - B)))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 6e-107)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_1);
	elseif (B <= 1.75e+16)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(F * t_0))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) / B) * Float64(-sqrt(Float64(F * Float64(A - B)))));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (B <= 6e-107)
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	elseif (B <= 1.75e+16)
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	else
		tmp = (sqrt(2.0) / B) * -sqrt((F * (A - B)));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 6e-107], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 1.75e+16], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(F * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / B), $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 5.9999999999999994e-107

   1. Initial program 24.5%

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

   3. Simplified 24.5%

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

      1. distribute-frac-neg 24.5%

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

   5. Applied egg-rr 33.2%

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

   6. Taylor expanded in A around -inf 17.0%

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

      1. unpow2 17.0%

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

   8. Simplified 17.0%

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

   if 5.9999999999999994e-107 < B < 1.75e16

   1. Initial program 32.8%

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

   3. Simplified 32.8%

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

   4. Taylor expanded in A around -inf 14.2%

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

      1. *-commutative 14.2%

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

   6. Simplified 14.2%

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

   if 1.75e16 < B

   1. Initial program 7.9%

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

   3. Simplified 7.9%

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

   4. Taylor expanded in C around 0 13.2%

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

      1. mul-1-neg 13.2%

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

      2. *-commutative 13.2%

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

      3. +-commutative 13.2%

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

      4. unpow2 13.2%

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

      5. unpow2 13.2%

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

      6. hypot-def 43.4%

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

   6. Simplified 43.4%

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

   7. Taylor expanded in A around 0 39.9%

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

      1. mul-1-neg 39.9%

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

      2. unsub-neg 39.9%

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

   9. Simplified 39.9%

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

4. Final simplification 22.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 6 \cdot 10^{-107}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 1.75 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{B} \cdot \left(-\sqrt{F \cdot \left(A - B\right)}\right)\\ \end{array} \]

Alternative 6: 42.3% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 2.9 \cdot 10^{-109}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 2 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot t_0\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= B 2.9e-109)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_1)
     (if (<= B 2e+16)
       (/ (- (sqrt (* 2.0 (* (* 2.0 A) (* F t_0))))) t_0)
       (* (sqrt (* F (- B))) (/ (- (sqrt 2.0)) B))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 2.9e-109) {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 2e+16) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	} else {
		tmp = sqrt((F * -B)) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (c * a))
    t_1 = (b * b) + ((-4.0d0) * (c * a))
    if (b <= 2.9d-109) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (a + (a + (0.5d0 * ((b * b) / a)))))))) / t_1
    else if (b <= 2d+16) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * (f * t_0)))) / t_0
    else
        tmp = sqrt((f * -b)) * (-sqrt(2.0d0) / b)
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 2.9e-109) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 2e+16) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	} else {
		tmp = Math.sqrt((F * -B)) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if B <= 2.9e-109:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1
	elif B <= 2e+16:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0
	else:
		tmp = math.sqrt((F * -B)) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 2.9e-109)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_1);
	elseif (B <= 2e+16)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(F * t_0))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(-B))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (B <= 2.9e-109)
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	elseif (B <= 2e+16)
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	else
		tmp = sqrt((F * -B)) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.9e-109], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 2e+16], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(F * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * (-B)), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 2.9e-109

   1. Initial program 24.5%

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

   3. Simplified 24.5%

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

      1. distribute-frac-neg 24.5%

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

   5. Applied egg-rr 33.2%

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

   6. Taylor expanded in A around -inf 17.0%

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

      1. unpow2 17.0%

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

   8. Simplified 17.0%

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

   if 2.9e-109 < B < 2e16

   1. Initial program 32.8%

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

   3. Simplified 32.8%

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

   4. Taylor expanded in A around -inf 14.2%

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

      1. *-commutative 14.2%

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

   6. Simplified 14.2%

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

   if 2e16 < B

   1. Initial program 7.9%

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

   3. Simplified 7.9%

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

   4. Taylor expanded in C around 0 13.2%

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

      1. mul-1-neg 13.2%

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

      2. *-commutative 13.2%

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

      3. +-commutative 13.2%

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

      4. unpow2 13.2%

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

      5. unpow2 13.2%

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

      6. hypot-def 43.4%

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

   6. Simplified 43.4%

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

   7. Taylor expanded in A around 0 40.4%

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

      1. mul-1-neg 40.4%

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

   9. Simplified 40.4%

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

4. Final simplification 22.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.9 \cdot 10^{-109}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 2 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 7: 42.3% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 1.72 \cdot 10^{-109}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 3.1 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot t_0\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(-B\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= B 1.72e-109)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_1)
     (if (<= B 3.1e+16)
       (/ (- (sqrt (* 2.0 (* (* 2.0 A) (* F t_0))))) t_0)
       (/ (* (sqrt (* F (- B))) (- (sqrt 2.0))) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 1.72e-109) {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 3.1e+16) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	} else {
		tmp = (sqrt((F * -B)) * -sqrt(2.0)) / B;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (c * a))
    t_1 = (b * b) + ((-4.0d0) * (c * a))
    if (b <= 1.72d-109) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (a + (a + (0.5d0 * ((b * b) / a)))))))) / t_1
    else if (b <= 3.1d+16) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * (f * t_0)))) / t_0
    else
        tmp = (sqrt((f * -b)) * -sqrt(2.0d0)) / b
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (B <= 1.72e-109) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (B <= 3.1e+16) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	} else {
		tmp = (Math.sqrt((F * -B)) * -Math.sqrt(2.0)) / B;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if B <= 1.72e-109:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1
	elif B <= 3.1e+16:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0
	else:
		tmp = (math.sqrt((F * -B)) * -math.sqrt(2.0)) / B
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (B <= 1.72e-109)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_1);
	elseif (B <= 3.1e+16)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(F * t_0))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(Float64(F * Float64(-B))) * Float64(-sqrt(2.0))) / B);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (B <= 1.72e-109)
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	elseif (B <= 3.1e+16)
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / t_0;
	else
		tmp = (sqrt((F * -B)) * -sqrt(2.0)) / B;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.72e-109], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 3.1e+16], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(F * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[N[(F * (-B)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 1.7200000000000001e-109

   1. Initial program 24.5%

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

   3. Simplified 24.5%

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

      1. distribute-frac-neg 24.5%

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

   5. Applied egg-rr 33.2%

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

   6. Taylor expanded in A around -inf 17.0%

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

      1. unpow2 17.0%

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

   8. Simplified 17.0%

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

   if 1.7200000000000001e-109 < B < 3.1e16

   1. Initial program 32.8%

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

   3. Simplified 32.8%

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

   4. Taylor expanded in A around -inf 14.2%

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

      1. *-commutative 14.2%

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

   6. Simplified 14.2%

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

   if 3.1e16 < B

   1. Initial program 7.9%

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

   3. Simplified 7.9%

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

   4. Taylor expanded in C around 0 13.2%

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

      1. mul-1-neg 13.2%

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

      2. *-commutative 13.2%

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

      3. +-commutative 13.2%

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

      4. unpow2 13.2%

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

      5. unpow2 13.2%

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

      6. hypot-def 43.4%

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

   6. Simplified 43.4%

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

   7. Taylor expanded in A around 0 40.4%

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

      1. mul-1-neg 40.4%

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

   9. Simplified 40.4%

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

       1. associate-*l/ 40.4%

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

   11. Applied egg-rr 40.4%

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

4. Final simplification 22.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.72 \cdot 10^{-109}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;B \leq 3.1 \cdot 10^{+16}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot \left(-B\right)} \cdot \left(-\sqrt{2}\right)}{B}\\ \end{array} \]

Alternative 8: 27.4% accurate, 4.3× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;A \leq -4 \cdot 10^{-90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;A \leq 2.75 \cdot 10^{-237}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(\left(C + A\right) - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)\right)}}{t_1}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= A -4e-90)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_1)
     (if (<= A 2.75e-237)
       (/ (- (sqrt (* 2.0 (* (* F t_0) (- (+ C A) B))))) t_0)
       (/
        (-
         (sqrt
          (*
           2.0
           (*
            t_1
            (*
             F
             (+ A (+ A (* -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C)))))))))
        t_1)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (A <= -4e-90) {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (A <= 2.75e-237) {
		tmp = -sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0;
	} else {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C)))))))) / t_1;
	}
	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 * (c * a))
    t_1 = (b * b) + ((-4.0d0) * (c * a))
    if (a <= (-4d-90)) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (a + (a + (0.5d0 * ((b * b) / a)))))))) / t_1
    else if (a <= 2.75d-237) then
        tmp = -sqrt((2.0d0 * ((f * t_0) * ((c + a) - b)))) / t_0
    else
        tmp = -sqrt((2.0d0 * (t_1 * (f * (a + (a + ((-0.5d0) * (((b * b) + ((a * a) - (a * a))) / c)))))))) / t_1
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (A <= -4e-90) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (A <= 2.75e-237) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0;
	} else {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C)))))))) / t_1;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (C * A))
	t_1 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if A <= -4e-90:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1
	elif A <= 2.75e-237:
		tmp = -math.sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0
	else:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C)))))))) / t_1
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(C * A)))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (A <= -4e-90)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_1);
	elseif (A <= 2.75e-237)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(Float64(C + A) - B))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(-0.5 * Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C))))))))) / t_1);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (C * A));
	t_1 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (A <= -4e-90)
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	elseif (A <= 2.75e-237)
		tmp = -sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0;
	else
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C)))))))) / t_1;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[A, -4e-90], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[A, 2.75e-237], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(N[(C + A), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if A < -3.99999999999999998e-90

   1. Initial program 21.0%

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

   3. Simplified 21.0%

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

      1. distribute-frac-neg 21.0%

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

   5. Applied egg-rr 33.5%

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

   6. Taylor expanded in A around -inf 30.6%

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

      1. unpow2 30.6%

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

   8. Simplified 30.6%

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

   if -3.99999999999999998e-90 < A < 2.74999999999999991e-237

   1. Initial program 43.3%

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

   3. Simplified 43.3%

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

   4. Taylor expanded in B around inf 14.9%

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

   if 2.74999999999999991e-237 < A

   1. Initial program 10.3%

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

   3. Simplified 10.3%

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

      1. distribute-frac-neg 10.3%

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

   5. Applied egg-rr 12.2%

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

   6. Taylor expanded in C around inf 7.6%

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

      1. associate--l+ 7.7%

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

      2. unpow2 7.7%

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

      3. unpow2 7.7%

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

      4. unpow2 7.7%

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

      5. mul-1-neg 7.7%

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

      6. mul-1-neg 7.7%

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

      7. sqr-neg 7.7%

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

   8. Simplified 7.7%

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

4. Final simplification 17.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -4 \cdot 10^{-90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;A \leq 2.75 \cdot 10^{-237}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(\left(C + A\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \end{array} \]

Alternative 9: 26.4% accurate, 4.6× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := B \cdot B + -4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;A \leq -4 \cdot 10^{-90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;A \leq 1.1 \cdot 10^{-232}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(\left(C + A\right) - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(F \cdot \left(C \cdot A\right)\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (+ (* B B) (* -4.0 (* C A)))))
   (if (<= A -4e-90)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (+ A (+ A (* 0.5 (/ (* B B) A))))))))) t_1)
     (if (<= A 1.1e-232)
       (/ (- (sqrt (* 2.0 (* (* F t_0) (- (+ C A) B))))) t_0)
       (/ (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* F (* 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 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (A <= -4e-90) {
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (A <= 1.1e-232) {
		tmp = -sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0;
	} else {
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * A)))))) / t_0;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (c * a))
    t_1 = (b * b) + ((-4.0d0) * (c * a))
    if (a <= (-4d-90)) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (a + (a + (0.5d0 * ((b * b) / a)))))))) / t_1
    else if (a <= 1.1d-232) then
        tmp = -sqrt((2.0d0 * ((f * t_0) * ((c + a) - b)))) / t_0
    else
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (f * (c * a)))))) / t_0
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = (B * B) + (-4.0 * (C * A));
	double tmp;
	if (A <= -4e-90) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	} else if (A <= 1.1e-232) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0;
	} else {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * 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 * (C * A))
	t_1 = (B * B) + (-4.0 * (C * A))
	tmp = 0
	if A <= -4e-90:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1
	elif A <= 1.1e-232:
		tmp = -math.sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0
	else:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (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(C * A)))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(C * A)))
	tmp = 0.0
	if (A <= -4e-90)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(A + Float64(A + Float64(0.5 * Float64(Float64(B * B) / A))))))))) / t_1);
	elseif (A <= 1.1e-232)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(Float64(C + A) - B))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(-4.0 * Float64(F * Float64(C * 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 * (C * A));
	t_1 = (B * B) + (-4.0 * (C * A));
	tmp = 0.0;
	if (A <= -4e-90)
		tmp = -sqrt((2.0 * (t_1 * (F * (A + (A + (0.5 * ((B * B) / A)))))))) / t_1;
	elseif (A <= 1.1e-232)
		tmp = -sqrt((2.0 * ((F * t_0) * ((C + A) - B)))) / t_0;
	else
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * 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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[A, -4e-90], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(A + N[(A + N[(0.5 * N[(N[(B * B), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[A, 1.1e-232], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(N[(C + A), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(-4.0 * N[(F * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if A < -3.99999999999999998e-90

   1. Initial program 21.0%

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

   3. Simplified 21.0%

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

      1. distribute-frac-neg 21.0%

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

   5. Applied egg-rr 33.5%

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

   6. Taylor expanded in A around -inf 30.6%

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

      1. unpow2 30.6%

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

   8. Simplified 30.6%

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

   if -3.99999999999999998e-90 < A < 1.10000000000000001e-232

   1. Initial program 43.3%

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

   3. Simplified 43.3%

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

   4. Taylor expanded in B around inf 14.9%

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

   if 1.10000000000000001e-232 < A

   1. Initial program 10.3%

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

   3. Simplified 10.3%

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

   4. Taylor expanded in A around -inf 3.9%

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

      1. *-commutative 3.9%

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

   6. Simplified 3.9%

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

   7. Taylor expanded in B around 0 5.9%

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

      1. associate-*r* 5.7%

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

      2. *-commutative 5.7%

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

   9. Simplified 5.7%

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

4. Final simplification 16.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -4 \cdot 10^{-90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(C \cdot A\right)\right) \cdot \left(F \cdot \left(A + \left(A + 0.5 \cdot \frac{B \cdot B}{A}\right)\right)\right)\right)}}{B \cdot B + -4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;A \leq 1.1 \cdot 10^{-232}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(\left(C + A\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(F \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \end{array} \]

Alternative 10: 26.9% accurate, 4.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ t_1 := F \cdot t_0\\ \mathbf{if}\;A \leq -4.2 \cdot 10^{-90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot t_1\right)}}{t_0}\\ \mathbf{elif}\;A \leq 1.32 \cdot 10^{-235}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(\left(C + A\right) - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(F \cdot \left(C \cdot A\right)\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* C A)))) (t_1 (* F t_0)))
   (if (<= A -4.2e-90)
     (/ (- (sqrt (* 2.0 (* (* 2.0 A) t_1)))) t_0)
     (if (<= A 1.32e-235)
       (/ (- (sqrt (* 2.0 (* t_1 (- (+ C A) B))))) t_0)
       (/ (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* F (* 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 * (C * A));
	double t_1 = F * t_0;
	double tmp;
	if (A <= -4.2e-90) {
		tmp = -sqrt((2.0 * ((2.0 * A) * t_1))) / t_0;
	} else if (A <= 1.32e-235) {
		tmp = -sqrt((2.0 * (t_1 * ((C + A) - B)))) / t_0;
	} else {
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * A)))))) / t_0;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (c * a))
    t_1 = f * t_0
    if (a <= (-4.2d-90)) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * t_1))) / t_0
    else if (a <= 1.32d-235) then
        tmp = -sqrt((2.0d0 * (t_1 * ((c + a) - b)))) / t_0
    else
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (f * (c * a)))))) / t_0
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (C * A));
	double t_1 = F * t_0;
	double tmp;
	if (A <= -4.2e-90) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * t_1))) / t_0;
	} else if (A <= 1.32e-235) {
		tmp = -Math.sqrt((2.0 * (t_1 * ((C + A) - B)))) / t_0;
	} else {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * 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 * (C * A))
	t_1 = F * t_0
	tmp = 0
	if A <= -4.2e-90:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * t_1))) / t_0
	elif A <= 1.32e-235:
		tmp = -math.sqrt((2.0 * (t_1 * ((C + A) - B)))) / t_0
	else:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (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(C * A)))
	t_1 = Float64(F * t_0)
	tmp = 0.0
	if (A <= -4.2e-90)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * t_1)))) / t_0);
	elseif (A <= 1.32e-235)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(Float64(C + A) - B))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(-4.0 * Float64(F * Float64(C * 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 * (C * A));
	t_1 = F * t_0;
	tmp = 0.0;
	if (A <= -4.2e-90)
		tmp = -sqrt((2.0 * ((2.0 * A) * t_1))) / t_0;
	elseif (A <= 1.32e-235)
		tmp = -sqrt((2.0 * (t_1 * ((C + A) - B)))) / t_0;
	else
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * 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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(F * t$95$0), $MachinePrecision]}, If[LessEqual[A, -4.2e-90], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * t$95$1), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[A, 1.32e-235], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(N[(C + A), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(-4.0 * N[(F * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if A < -4.1999999999999998e-90

   1. Initial program 21.0%

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

   3. Simplified 21.0%

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

   4. Taylor expanded in A around -inf 29.4%

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

      1. *-commutative 29.4%

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

   6. Simplified 29.4%

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

   if -4.1999999999999998e-90 < A < 1.32e-235

   1. Initial program 43.3%

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

   3. Simplified 43.3%

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

   4. Taylor expanded in B around inf 14.9%

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

   if 1.32e-235 < A

   1. Initial program 10.3%

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

   3. Simplified 10.3%

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

   4. Taylor expanded in A around -inf 3.9%

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

      1. *-commutative 3.9%

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

   6. Simplified 3.9%

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

   7. Taylor expanded in B around 0 5.9%

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

      1. associate-*r* 5.7%

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

      2. *-commutative 5.7%

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

   9. Simplified 5.7%

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

4. Final simplification 15.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -4.2 \cdot 10^{-90}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{elif}\;A \leq 1.32 \cdot 10^{-235}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right) \cdot \left(\left(C + A\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(F \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \end{array} \]

Alternative 11: 28.9% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(C \cdot A\right)\\ \mathbf{if}\;B \leq 2.8 \cdot 10^{+116}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot t_0\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

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 (* C A)))))
   (if (<= B 2.8e+116)
     (/ (- (sqrt (* 2.0 (* (* 2.0 A) (* F t_0))))) 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 * (C * A));
	double tmp;
	if (B <= 2.8e+116) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / 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 * (c * a))
    if (b <= 2.8d+116) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * (f * t_0)))) / 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 * (C * A));
	double tmp;
	if (B <= 2.8e+116) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / 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 * (C * A))
	tmp = 0
	if B <= 2.8e+116:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / 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(C * A)))
	tmp = 0.0
	if (B <= 2.8e+116)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(F * t_0))))) / 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 * (C * A));
	tmp = 0.0;
	if (B <= 2.8e+116)
		tmp = -sqrt((2.0 * ((2.0 * A) * (F * t_0)))) / 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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 2.8e+116], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(F * t$95$0), $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 < 2.80000000000000004e116

   1. Initial program 24.6%

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

   3. Simplified 24.6%

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

   4. Taylor expanded in A around -inf 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%

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

   if 2.80000000000000004e116 < B

   1. Initial program 3.2%

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

   3. Simplified 3.2%

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

   4. Taylor expanded in A around -inf 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 2.8%

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

   8. Taylor expanded in B around inf 6.4%

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

      1. associate-*r/ 6.4%

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

      2. *-commutative 6.4%

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

      3. *-rgt-identity 6.4%

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

   10. Simplified 6.4%

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

4. Final simplification 14.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 2.8 \cdot 10^{+116}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(F \cdot \left(B \cdot B - 4 \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 12: 22.5% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 4.3 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (if (<= B 4.3e-14)
   (/
    (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* A (* C F)))))))
    (- (* B B) (* 4.0 (* C A))))
   (* -2.0 (/ (sqrt (* A F)) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 4.3e-14) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (sqrt((A * F)) / B);
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: tmp
    if (b <= 4.3d-14) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (a * (c * f)))))) / ((b * b) - (4.0d0 * (c * a)))
    else
        tmp = (-2.0d0) * (sqrt((a * f)) / b)
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 4.3e-14) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 4.3e-14:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)))
	else:
		tmp = -2.0 * (math.sqrt((A * F)) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 4.3e-14)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(-4.0 * Float64(A * Float64(C * F))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(C * A))));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 4.3e-14)
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - (4.0 * (C * A)));
	else
		tmp = -2.0 * (sqrt((A * F)) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := If[LessEqual[B, 4.3e-14], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(-4.0 * N[(A * N[(C * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 4.29999999999999998e-14

   1. Initial program 25.3%

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

   3. Simplified 25.3%

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

   4. Taylor expanded in A around -inf 15.0%

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

      1. *-commutative 15.0%

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

   6. Simplified 15.0%

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

   7. Taylor expanded in B around 0 11.0%

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

   if 4.29999999999999998e-14 < B

   1. Initial program 10.3%

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

   3. Simplified 10.3%

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

   4. Taylor expanded in A around -inf 8.5%

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

      1. *-commutative 8.5%

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

   6. Simplified 8.5%

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

   7. Taylor expanded in B around inf 5.7%

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

   8. Taylor expanded in B around inf 7.7%

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

      1. associate-*r/ 7.7%

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

      2. *-commutative 7.7%

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

      3. *-rgt-identity 7.7%

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

   10. Simplified 7.7%

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

4. Final simplification 10.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 4.3 \cdot 10^{-14}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 13: 26.6% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 210000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(F \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (if (<= B 210000000.0)
   (/
    (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* F (* C A)))))))
    (- (* B B) (* 4.0 (* C A))))
   (* -2.0 (/ (sqrt (* A F)) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 210000000.0) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * A)))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (sqrt((A * F)) / B);
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: tmp
    if (b <= 210000000.0d0) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (f * (c * a)))))) / ((b * b) - (4.0d0 * (c * a)))
    else
        tmp = (-2.0d0) * (sqrt((a * f)) / b)
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 210000000.0) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * A)))))) / ((B * B) - (4.0 * (C * A)));
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 210000000.0:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * A)))))) / ((B * B) - (4.0 * (C * A)))
	else:
		tmp = -2.0 * (math.sqrt((A * F)) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 210000000.0)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(2.0 * A) * Float64(-4.0 * Float64(F * Float64(C * A))))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(C * A))));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 210000000.0)
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (F * (C * A)))))) / ((B * B) - (4.0 * (C * A)));
	else
		tmp = -2.0 * (sqrt((A * F)) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := If[LessEqual[B, 210000000.0], N[((-N[Sqrt[N[(2.0 * N[(N[(2.0 * A), $MachinePrecision] * N[(-4.0 * N[(F * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 2.1e8

   1. Initial program 25.3%

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

   3. Simplified 25.3%

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

   4. Taylor expanded in A around -inf 15.7%

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

      1. *-commutative 15.7%

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

   6. Simplified 15.7%

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

   7. Taylor expanded in B around 0 11.2%

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

      1. associate-*r* 13.6%

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

      2. *-commutative 13.6%

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

   9. Simplified 13.6%

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

   if 2.1e8 < B

   1. Initial program 9.4%

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

   3. Simplified 9.4%

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

   4. Taylor expanded in A around -inf 5.8%

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

      1. *-commutative 5.8%

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

   6. Simplified 5.8%

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

   7. Taylor expanded in B around inf 5.7%

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

   8. Taylor expanded in B around inf 7.9%

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

      1. associate-*r/ 7.9%

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

      2. *-commutative 7.9%

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

      3. *-rgt-identity 7.9%

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

   10. Simplified 7.9%

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

4. Final simplification 12.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 210000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(F \cdot \left(C \cdot A\right)\right)\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \end{array} \]

Alternative 14: 18.9% accurate, 5.1× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;A \leq -1.12 \cdot 10^{+111}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(C \cdot F\right) \cdot \left(\left(A \cdot A\right) \cdot -8\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (if (<= A -1.12e+111)
   (* -2.0 (/ (sqrt (* A F)) B))
   (/
    (- (sqrt (* 2.0 (* (* C F) (* (* A A) -8.0)))))
    (- (* B B) (* 4.0 (* C A))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (A <= -1.12e+111) {
		tmp = -2.0 * (sqrt((A * F)) / B);
	} else {
		tmp = -sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (C * A)));
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: tmp
    if (a <= (-1.12d+111)) then
        tmp = (-2.0d0) * (sqrt((a * f)) / b)
    else
        tmp = -sqrt((2.0d0 * ((c * f) * ((a * a) * (-8.0d0))))) / ((b * b) - (4.0d0 * (c * a)))
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double tmp;
	if (A <= -1.12e+111) {
		tmp = -2.0 * (Math.sqrt((A * F)) / B);
	} else {
		tmp = -Math.sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (C * A)));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if A <= -1.12e+111:
		tmp = -2.0 * (math.sqrt((A * F)) / B)
	else:
		tmp = -math.sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (C * A)))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (A <= -1.12e+111)
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) / B));
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(C * F) * Float64(Float64(A * A) * -8.0))))) / Float64(Float64(B * B) - Float64(4.0 * Float64(C * A))));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (A <= -1.12e+111)
		tmp = -2.0 * (sqrt((A * F)) / B);
	else
		tmp = -sqrt((2.0 * ((C * F) * ((A * A) * -8.0)))) / ((B * B) - (4.0 * (C * A)));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := If[LessEqual[A, -1.12e+111], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision], 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[(C * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if A < -1.11999999999999995e111

   1. Initial program 10.9%

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

   3. Simplified 10.9%

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

   4. Taylor expanded in A around -inf 31.1%

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

      1. *-commutative 31.1%

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

   6. Simplified 31.1%

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

   7. Taylor expanded in B around inf 6.2%

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

   8. Taylor expanded in B around inf 9.0%

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

      1. associate-*r/ 9.0%

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

      2. *-commutative 9.0%

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

      3. *-rgt-identity 9.0%

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

   10. Simplified 9.0%

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

   if -1.11999999999999995e111 < A

   1. Initial program 23.6%

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

   3. Simplified 23.6%

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

   4. Taylor expanded in A around -inf 9.8%

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

      1. *-commutative 9.8%

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

   6. Simplified 9.8%

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

   7. Taylor expanded in B around 0 6.5%

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

      1. associate-*r* 6.4%

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

      2. unpow2 6.4%

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

      3. *-commutative 6.4%

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

   9. Simplified 6.4%

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

4. Final simplification 6.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -1.12 \cdot 10^{+111}:\\ \;\;\;\;-2 \cdot \frac{\sqrt{A \cdot F}}{B}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(C \cdot F\right) \cdot \left(\left(A \cdot A\right) \cdot -8\right)\right)}}{B \cdot B - 4 \cdot \left(C \cdot A\right)}\\ \end{array} \]

Alternative 15: 9.2% 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 21.5%

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

3. Simplified 21.5%

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

4. Taylor expanded in A around -inf 13.4%

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

   1. *-commutative 13.4%

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

6. Simplified 13.4%

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

7. Taylor expanded in B around inf 2.4%

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

8. Taylor expanded in B around inf 2.9%

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

   1. associate-*r/ 2.9%

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

   2. *-commutative 2.9%

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

   3. *-rgt-identity 2.9%

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

10. Simplified 2.9%

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

11. Final simplification 2.9%

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

Reproduce

herbie shell --seed 2023240 
(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))))