ABCF->ab-angle b

Percentage Accurate: 19.1% → 46.3%

Time: 21.0s

Alternatives: 12

Speedup: 5.0×

Specification

Math

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

FPCore

(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (pow B 2.0) (* (* 4.0 A) C))))
   (/
    (-
     (sqrt
      (*
       (* 2.0 (* t_0 F))
       (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0)))))))
    t_0)))

C

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

Fortran

real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b ** 2.0d0) - ((4.0d0 * a) * c)
    code = -sqrt(((2.0d0 * (t_0 * f)) * ((a + c) - sqrt((((a - c) ** 2.0d0) + (b ** 2.0d0)))))) / t_0
end function

Java

public static double code(double A, double B, double C, double F) {
	double t_0 = Math.pow(B, 2.0) - ((4.0 * A) * C);
	return -Math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - Math.sqrt((Math.pow((A - C), 2.0) + Math.pow(B, 2.0)))))) / t_0;
}

Python

def code(A, B, C, F):
	t_0 = math.pow(B, 2.0) - ((4.0 * A) * C)
	return -math.sqrt(((2.0 * (t_0 * F)) * ((A + C) - math.sqrt((math.pow((A - C), 2.0) + math.pow(B, 2.0)))))) / t_0

Julia

function code(A, B, C, F)
	t_0 = Float64((B ^ 2.0) - Float64(Float64(4.0 * A) * C))
	return Float64(Float64(-sqrt(Float64(Float64(2.0 * Float64(t_0 * F)) * Float64(Float64(A + C) - sqrt(Float64((Float64(A - C) ^ 2.0) + (B ^ 2.0))))))) / t_0)
end

MATLAB

function tmp = code(A, B, C, F)
	t_0 = (B ^ 2.0) - ((4.0 * A) * C);
	tmp = -sqrt(((2.0 * (t_0 * F)) * ((A + C) - sqrt((((A - C) ^ 2.0) + (B ^ 2.0)))))) / t_0;
end

Wolfram

code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[Power[B, 2.0], $MachinePrecision] - N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]), $MachinePrecision]}, N[((-N[Sqrt[N[(N[(2.0 * N[(t$95$0 * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - N[Sqrt[N[(N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[B, 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]

Initial Program: 19.1% 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: 46.3% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 20500000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (+ (* B B) (* -4.0 (* A C)))))
   (if (<= B 20500000000.0)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (/ (* (sqrt 2.0) (- (sqrt (* F (- A (hypot A B)))))) B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 20500000000.0) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = (sqrt(2.0) * -sqrt((F * (A - hypot(A, B))))) / B;
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 20500000000.0) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = (Math.sqrt(2.0) * -Math.sqrt((F * (A - Math.hypot(A, B))))) / B;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 20500000000.0:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	else:
		tmp = (math.sqrt(2.0) * -math.sqrt((F * (A - math.hypot(A, B))))) / B
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 20500000000.0)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(Float64(sqrt(2.0) * Float64(-sqrt(Float64(F * Float64(A - hypot(A, B)))))) / B);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 20500000000.0)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	else
		tmp = (sqrt(2.0) * -sqrt((F * (A - hypot(A, B))))) / B;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 20500000000.0], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] * (-N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision])), $MachinePrecision] / B), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 2.05e10

   1. Initial program 17.3%

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

   3. Simplified 17.3%

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

   4. Taylor expanded in A around -inf 15.1%

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

      1. *-commutative 15.1%

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

   6. Simplified 15.1%

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

      1. distribute-frac-neg 15.1%

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

      2. associate-*l* 14.6%

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

      3. cancel-sign-sub-inv 14.6%

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

      4. metadata-eval 14.6%

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

      5. cancel-sign-sub-inv 14.6%

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

      6. metadata-eval 14.6%

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

   8. Applied egg-rr 14.6%

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

   if 2.05e10 < B

   1. Initial program 20.4%

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

   3. Simplified 20.4%

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

   4. Taylor expanded in C around 0 23.9%

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

      1. mul-1-neg 23.9%

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

      2. *-commutative 23.9%

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

      3. +-commutative 23.9%

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

      4. unpow2 23.9%

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

      5. unpow2 23.9%

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

      6. hypot-def 50.6%

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

   6. Simplified 50.6%

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

      1. associate-*l/ 50.7%

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

   8. Applied egg-rr 50.7%

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

4. Final simplification 22.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 20500000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2} \cdot \left(-\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)}\right)}{B}\\ \end{array} \]

Alternative 2: 46.3% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 30500000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (+ (* B B) (* -4.0 (* A C)))))
   (if (<= B 30500000000.0)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (* (sqrt (* F (- A (hypot A B)))) (/ (- (sqrt 2.0)) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 30500000000.0) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 30500000000.0) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = Math.sqrt((F * (A - Math.hypot(A, B)))) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 30500000000.0:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	else:
		tmp = math.sqrt((F * (A - math.hypot(A, B)))) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 30500000000.0)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 30500000000.0)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	else
		tmp = sqrt((F * (A - hypot(A, B)))) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 30500000000.0], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 3.05e10

   1. Initial program 17.3%

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

   3. Simplified 17.3%

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

   4. Taylor expanded in A around -inf 15.1%

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

      1. *-commutative 15.1%

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

   6. Simplified 15.1%

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

      1. distribute-frac-neg 15.1%

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

      2. associate-*l* 14.6%

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

      3. cancel-sign-sub-inv 14.6%

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

      4. metadata-eval 14.6%

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

      5. cancel-sign-sub-inv 14.6%

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

      6. metadata-eval 14.6%

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

   8. Applied egg-rr 14.6%

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

   if 3.05e10 < B

   1. Initial program 20.4%

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

   3. Simplified 20.4%

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

   4. Taylor expanded in C around 0 23.9%

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

      1. mul-1-neg 23.9%

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

      2. *-commutative 23.9%

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

      3. +-commutative 23.9%

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

      4. unpow2 23.9%

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

      5. unpow2 23.9%

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

      6. hypot-def 50.6%

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

   6. Simplified 50.6%

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

4. Final simplification 22.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 30500000000:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 3: 43.0% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 3.5 \cdot 10^{+40}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

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 (* A C)))))
   (if (<= B 3.5e+40)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (* (sqrt (* F (- A B))) (/ (- (sqrt 2.0)) B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 3.5e+40) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = sqrt((F * (A - 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) :: tmp
    t_0 = (b * b) + ((-4.0d0) * (a * c))
    if (b <= 3.5d+40) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / t_0
    else
        tmp = sqrt((f * (a - 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 * (A * C));
	double tmp;
	if (B <= 3.5e+40) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = Math.sqrt((F * (A - B))) * (-Math.sqrt(2.0) / B);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 3.5e+40:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	else:
		tmp = math.sqrt((F * (A - B))) * (-math.sqrt(2.0) / B)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 3.5e+40)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - B))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 3.5e+40)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	else
		tmp = sqrt((F * (A - B))) * (-sqrt(2.0) / B);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.5e+40], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - B), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 3.4999999999999999e40

   1. Initial program 17.4%

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

   3. Simplified 17.4%

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

   4. Taylor expanded in A around -inf 15.3%

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

      1. *-commutative 15.3%

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

   6. Simplified 15.3%

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

      1. distribute-frac-neg 15.3%

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

      2. associate-*l* 14.8%

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

      3. cancel-sign-sub-inv 14.8%

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

      4. metadata-eval 14.8%

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

      5. cancel-sign-sub-inv 14.8%

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

      6. metadata-eval 14.8%

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

   8. Applied egg-rr 14.8%

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

   if 3.4999999999999999e40 < B

   1. Initial program 20.1%

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

   3. Simplified 20.1%

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

   4. Taylor expanded in C around 0 23.6%

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

      1. mul-1-neg 23.6%

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

      2. *-commutative 23.6%

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

      3. +-commutative 23.6%

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

      4. unpow2 23.6%

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

      5. unpow2 23.6%

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

      6. hypot-def 52.3%

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

   6. Simplified 52.3%

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

   7. Taylor expanded in A around 0 50.4%

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

      1. mul-1-neg 50.4%

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

      2. unsub-neg 50.4%

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

   9. Simplified 50.4%

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

4. Final simplification 21.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.5 \cdot 10^{+40}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 4: 42.6% accurate, 3.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 1.15 \cdot 10^{+41}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

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 (* A C)))))
   (if (<= B 1.15e+41)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) 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 * (A * C));
	double tmp;
	if (B <= 1.15e+41) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / 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) :: tmp
    t_0 = (b * b) + ((-4.0d0) * (a * c))
    if (b <= 1.15d+41) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / 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 * (A * C));
	double tmp;
	if (B <= 1.15e+41) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / 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 * (A * C))
	tmp = 0
	if B <= 1.15e+41:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / 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(A * C)))
	tmp = 0.0
	if (B <= 1.15e+41)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(-B))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 1.15e+41)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / 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[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 1.15e+41], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * (-B)), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 1.1499999999999999e41

   1. Initial program 17.4%

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

   3. Simplified 17.4%

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

   4. Taylor expanded in A around -inf 15.3%

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

      1. *-commutative 15.3%

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

   6. Simplified 15.3%

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

      1. distribute-frac-neg 15.3%

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

      2. associate-*l* 14.8%

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

      3. cancel-sign-sub-inv 14.8%

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

      4. metadata-eval 14.8%

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

      5. cancel-sign-sub-inv 14.8%

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

      6. metadata-eval 14.8%

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

   8. Applied egg-rr 14.8%

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

   if 1.1499999999999999e41 < B

   1. Initial program 20.1%

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

   3. Simplified 20.1%

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

   4. Taylor expanded in C around 0 23.6%

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

      1. mul-1-neg 23.6%

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

      2. *-commutative 23.6%

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

      3. +-commutative 23.6%

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

      4. unpow2 23.6%

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

      5. unpow2 23.6%

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

      6. hypot-def 52.3%

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

   6. Simplified 52.3%

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

   7. Taylor expanded in A around 0 50.6%

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

      1. mul-1-neg 50.6%

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

   9. Simplified 50.6%

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

4. Final simplification 21.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.15 \cdot 10^{+41}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(-B\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 5: 30.1% accurate, 4.5× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - \left(A \cdot C\right) \cdot 4\\ t_1 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 3.9 \cdot 10^{+43}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 3.9 \cdot 10^{+144}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(C + \left(-0.5 \cdot \frac{C \cdot C}{B} - B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* (* A C) 4.0))) (t_1 (+ (* B B) (* -4.0 (* A C)))))
   (if (<= B 3.9e+43)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (* 2.0 A)))))) t_1)
     (if (<= B 3.9e+144)
       (-
        (/
         (sqrt (* 2.0 (* (* F t_0) (+ C (- (* -0.5 (/ (* C C) B)) B)))))
         t_0))
       (* -2.0 (* (sqrt (* A F)) (/ 1.0 B)))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - ((A * C) * 4.0);
	double t_1 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 3.9e+43) {
		tmp = -sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	} else if (B <= 3.9e+144) {
		tmp = -(sqrt((2.0 * ((F * t_0) * (C + ((-0.5 * ((C * C) / B)) - B))))) / t_0);
	} else {
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / 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) - ((a * c) * 4.0d0)
    t_1 = (b * b) + ((-4.0d0) * (a * c))
    if (b <= 3.9d+43) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (2.0d0 * a))))) / t_1
    else if (b <= 3.9d+144) then
        tmp = -(sqrt((2.0d0 * ((f * t_0) * (c + (((-0.5d0) * ((c * c) / b)) - b))))) / t_0)
    else
        tmp = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / 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) - ((A * C) * 4.0);
	double t_1 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 3.9e+43) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	} else if (B <= 3.9e+144) {
		tmp = -(Math.sqrt((2.0 * ((F * t_0) * (C + ((-0.5 * ((C * C) / B)) - B))))) / t_0);
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) * (1.0 / B));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - ((A * C) * 4.0)
	t_1 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 3.9e+43:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1
	elif B <= 3.9e+144:
		tmp = -(math.sqrt((2.0 * ((F * t_0) * (C + ((-0.5 * ((C * C) / B)) - B))))) / t_0)
	else:
		tmp = -2.0 * (math.sqrt((A * F)) * (1.0 / B))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(Float64(A * C) * 4.0))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 3.9e+43)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(2.0 * A)))))) / t_1);
	elseif (B <= 3.9e+144)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(C + Float64(Float64(-0.5 * Float64(Float64(C * C) / B)) - B))))) / t_0));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / B)));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - ((A * C) * 4.0);
	t_1 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 3.9e+43)
		tmp = -sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	elseif (B <= 3.9e+144)
		tmp = -(sqrt((2.0 * ((F * t_0) * (C + ((-0.5 * ((C * C) / B)) - B))))) / t_0);
	else
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(N[(A * C), $MachinePrecision] * 4.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.9e+43], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 3.9e+144], (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(C + N[(N[(-0.5 * N[(N[(C * C), $MachinePrecision] / B), $MachinePrecision]), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 3.9000000000000001e43

   1. Initial program 17.4%

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

   3. Simplified 17.4%

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

   4. Taylor expanded in A around -inf 15.3%

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

      1. *-commutative 15.3%

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

   6. Simplified 15.3%

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

      1. distribute-frac-neg 15.3%

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

      2. associate-*l* 14.8%

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

      3. cancel-sign-sub-inv 14.8%

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

      4. metadata-eval 14.8%

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

      5. cancel-sign-sub-inv 14.8%

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

      6. metadata-eval 14.8%

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

   8. Applied egg-rr 14.8%

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

   if 3.9000000000000001e43 < B < 3.90000000000000018e144

   1. Initial program 53.8%

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

   3. Simplified 53.8%

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

   4. Taylor expanded in A around 0 53.8%

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

      1. +-commutative 53.8%

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

      2. unpow2 53.8%

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

      3. unpow2 53.8%

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

      4. hypot-def 53.6%

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

   6. Simplified 53.6%

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

   7. Taylor expanded in C around 0 49.2%

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

      1. mul-1-neg 49.2%

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

      2. unsub-neg 49.2%

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

      3. unpow2 49.2%

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

   9. Simplified 49.2%

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

   if 3.90000000000000018e144 < B

   1. Initial program 0.1%

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

   3. Simplified 0.1%

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

   4. Taylor expanded in A around -inf 0.0%

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

      1. *-commutative 0.0%

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

   6. Simplified 0.0%

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

   7. Taylor expanded in B around inf 5.6%

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

      1. *-commutative 5.6%

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

   9. Simplified 5.6%

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

4. Final simplification 16.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.9 \cdot 10^{+43}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 3.9 \cdot 10^{+144}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - \left(A \cdot C\right) \cdot 4\right)\right) \cdot \left(C + \left(-0.5 \cdot \frac{C \cdot C}{B} - B\right)\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \]

Alternative 6: 30.1% accurate, 4.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - \left(A \cdot C\right) \cdot 4\\ t_1 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 9.5 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 1.25 \cdot 10^{+146}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* (* A C) 4.0))) (t_1 (+ (* B B) (* -4.0 (* A C)))))
   (if (<= B 9.5e+42)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (* 2.0 A)))))) t_1)
     (if (<= B 1.25e+146)
       (/ (- (sqrt (* 2.0 (* (* F t_0) (- (+ A C) B))))) t_0)
       (* -2.0 (* (sqrt (* A F)) (/ 1.0 B)))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - ((A * C) * 4.0);
	double t_1 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 9.5e+42) {
		tmp = -sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	} else if (B <= 1.25e+146) {
		tmp = -sqrt((2.0 * ((F * t_0) * ((A + C) - B)))) / t_0;
	} else {
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / 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) - ((a * c) * 4.0d0)
    t_1 = (b * b) + ((-4.0d0) * (a * c))
    if (b <= 9.5d+42) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (2.0d0 * a))))) / t_1
    else if (b <= 1.25d+146) then
        tmp = -sqrt((2.0d0 * ((f * t_0) * ((a + c) - b)))) / t_0
    else
        tmp = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / 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) - ((A * C) * 4.0);
	double t_1 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 9.5e+42) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	} else if (B <= 1.25e+146) {
		tmp = -Math.sqrt((2.0 * ((F * t_0) * ((A + C) - B)))) / t_0;
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) * (1.0 / B));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - ((A * C) * 4.0)
	t_1 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 9.5e+42:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1
	elif B <= 1.25e+146:
		tmp = -math.sqrt((2.0 * ((F * t_0) * ((A + C) - B)))) / t_0
	else:
		tmp = -2.0 * (math.sqrt((A * F)) * (1.0 / B))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(Float64(A * C) * 4.0))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 9.5e+42)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(2.0 * A)))))) / t_1);
	elseif (B <= 1.25e+146)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(Float64(A + C) - B))))) / t_0);
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / B)));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - ((A * C) * 4.0);
	t_1 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 9.5e+42)
		tmp = -sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	elseif (B <= 1.25e+146)
		tmp = -sqrt((2.0 * ((F * t_0) * ((A + C) - B)))) / t_0;
	else
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(N[(A * C), $MachinePrecision] * 4.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 9.5e+42], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 1.25e+146], N[((-N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 9.50000000000000019e42

   1. Initial program 17.4%

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

   3. Simplified 17.4%

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

   4. Taylor expanded in A around -inf 15.3%

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

      1. *-commutative 15.3%

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

   6. Simplified 15.3%

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

      1. distribute-frac-neg 15.3%

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

      2. associate-*l* 14.8%

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

      3. cancel-sign-sub-inv 14.8%

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

      4. metadata-eval 14.8%

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

      5. cancel-sign-sub-inv 14.8%

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

      6. metadata-eval 14.8%

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

   8. Applied egg-rr 14.8%

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

   if 9.50000000000000019e42 < B < 1.25e146

   1. Initial program 53.8%

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

   3. Simplified 53.8%

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

   4. Taylor expanded in B around inf 49.0%

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

   if 1.25e146 < B

   1. Initial program 0.1%

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

   3. Simplified 0.1%

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

   4. Taylor expanded in A around -inf 0.0%

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

      1. *-commutative 0.0%

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

   6. Simplified 0.0%

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

   7. Taylor expanded in B around inf 5.6%

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

      1. *-commutative 5.6%

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

   9. Simplified 5.6%

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

4. Final simplification 16.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 9.5 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 1.25 \cdot 10^{+146}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - \left(A \cdot C\right) \cdot 4\right)\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \]

Alternative 7: 29.9% accurate, 4.8× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - \left(A \cdot C\right) \cdot 4\\ t_1 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 9.5 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_1}\\ \mathbf{elif}\;B \leq 1.1 \cdot 10^{+146}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot t_0\right) \cdot \left(C - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* (* A C) 4.0))) (t_1 (+ (* B B) (* -4.0 (* A C)))))
   (if (<= B 9.5e+42)
     (/ (- (sqrt (* 2.0 (* t_1 (* F (* 2.0 A)))))) t_1)
     (if (<= B 1.1e+146)
       (- (/ (sqrt (* 2.0 (* (* F t_0) (- C B)))) t_0))
       (* -2.0 (* (sqrt (* A F)) (/ 1.0 B)))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - ((A * C) * 4.0);
	double t_1 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 9.5e+42) {
		tmp = -sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	} else if (B <= 1.1e+146) {
		tmp = -(sqrt((2.0 * ((F * t_0) * (C - B)))) / t_0);
	} else {
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / 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) - ((a * c) * 4.0d0)
    t_1 = (b * b) + ((-4.0d0) * (a * c))
    if (b <= 9.5d+42) then
        tmp = -sqrt((2.0d0 * (t_1 * (f * (2.0d0 * a))))) / t_1
    else if (b <= 1.1d+146) then
        tmp = -(sqrt((2.0d0 * ((f * t_0) * (c - b)))) / t_0)
    else
        tmp = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / 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) - ((A * C) * 4.0);
	double t_1 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 9.5e+42) {
		tmp = -Math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	} else if (B <= 1.1e+146) {
		tmp = -(Math.sqrt((2.0 * ((F * t_0) * (C - B)))) / t_0);
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) * (1.0 / B));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - ((A * C) * 4.0)
	t_1 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 9.5e+42:
		tmp = -math.sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1
	elif B <= 1.1e+146:
		tmp = -(math.sqrt((2.0 * ((F * t_0) * (C - B)))) / t_0)
	else:
		tmp = -2.0 * (math.sqrt((A * F)) * (1.0 / B))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(Float64(A * C) * 4.0))
	t_1 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 9.5e+42)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(F * Float64(2.0 * A)))))) / t_1);
	elseif (B <= 1.1e+146)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(Float64(F * t_0) * Float64(C - B)))) / t_0));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / B)));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - ((A * C) * 4.0);
	t_1 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 9.5e+42)
		tmp = -sqrt((2.0 * (t_1 * (F * (2.0 * A))))) / t_1;
	elseif (B <= 1.1e+146)
		tmp = -(sqrt((2.0 * ((F * t_0) * (C - B)))) / t_0);
	else
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(N[(A * C), $MachinePrecision] * 4.0), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 9.5e+42], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$1), $MachinePrecision], If[LessEqual[B, 1.1e+146], (-N[(N[Sqrt[N[(2.0 * N[(N[(F * t$95$0), $MachinePrecision] * N[(C - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if B < 9.50000000000000019e42

   1. Initial program 17.4%

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

   3. Simplified 17.4%

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

   4. Taylor expanded in A around -inf 15.3%

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

      1. *-commutative 15.3%

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

   6. Simplified 15.3%

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

      1. distribute-frac-neg 15.3%

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

      2. associate-*l* 14.8%

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

      3. cancel-sign-sub-inv 14.8%

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

      4. metadata-eval 14.8%

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

      5. cancel-sign-sub-inv 14.8%

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

      6. metadata-eval 14.8%

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

   8. Applied egg-rr 14.8%

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

   if 9.50000000000000019e42 < B < 1.0999999999999999e146

   1. Initial program 53.8%

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

   3. Simplified 53.8%

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

   4. Taylor expanded in A around 0 53.8%

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

      1. +-commutative 53.8%

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

      2. unpow2 53.8%

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

      3. unpow2 53.8%

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

      4. hypot-def 53.6%

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

   6. Simplified 53.6%

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

   7. Taylor expanded in C around 0 49.4%

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

      1. mul-1-neg 49.4%

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

      2. unsub-neg 49.4%

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

   9. Simplified 49.4%

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

   if 1.0999999999999999e146 < B

   1. Initial program 0.1%

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

   3. Simplified 0.1%

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

   4. Taylor expanded in A around -inf 0.0%

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

      1. *-commutative 0.0%

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

   6. Simplified 0.0%

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

   7. Taylor expanded in B around inf 5.6%

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

      1. *-commutative 5.6%

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

   9. Simplified 5.6%

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

4. Final simplification 16.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 9.5 \cdot 10^{+42}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;B \leq 1.1 \cdot 10^{+146}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(F \cdot \left(B \cdot B - \left(A \cdot C\right) \cdot 4\right)\right) \cdot \left(C - B\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \]

Alternative 8: 28.6% accurate, 4.9× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B + -4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 3.2 \cdot 10^{+43}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_0 \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (+ (* B B) (* -4.0 (* A C)))))
   (if (<= B 3.2e+43)
     (/ (- (sqrt (* 2.0 (* t_0 (* F (* 2.0 A)))))) t_0)
     (* -2.0 (* (sqrt (* A F)) (/ 1.0 B))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) + (-4.0 * (A * C));
	double tmp;
	if (B <= 3.2e+43) {
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = (b * b) + ((-4.0d0) * (a * c))
    if (b <= 3.2d+43) then
        tmp = -sqrt((2.0d0 * (t_0 * (f * (2.0d0 * a))))) / t_0
    else
        tmp = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / 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 * (A * C));
	double tmp;
	if (B <= 3.2e+43) {
		tmp = -Math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) * (1.0 / B));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) + (-4.0 * (A * C))
	tmp = 0
	if B <= 3.2e+43:
		tmp = -math.sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0
	else:
		tmp = -2.0 * (math.sqrt((A * F)) * (1.0 / B))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) + Float64(-4.0 * Float64(A * C)))
	tmp = 0.0
	if (B <= 3.2e+43)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_0 * Float64(F * Float64(2.0 * A)))))) / t_0);
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / B)));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) + (-4.0 * (A * C));
	tmp = 0.0;
	if (B <= 3.2e+43)
		tmp = -sqrt((2.0 * (t_0 * (F * (2.0 * A))))) / t_0;
	else
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[B, 3.2e+43], N[((-N[Sqrt[N[(2.0 * N[(t$95$0 * N[(F * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 3.20000000000000014e43

   1. Initial program 17.4%

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

   3. Simplified 17.4%

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

   4. Taylor expanded in A around -inf 15.3%

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

      1. *-commutative 15.3%

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

   6. Simplified 15.3%

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

      1. distribute-frac-neg 15.3%

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

      2. associate-*l* 14.8%

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

      3. cancel-sign-sub-inv 14.8%

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

      4. metadata-eval 14.8%

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

      5. cancel-sign-sub-inv 14.8%

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

      6. metadata-eval 14.8%

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

   8. Applied egg-rr 14.8%

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

   if 3.20000000000000014e43 < B

   1. Initial program 20.1%

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

   3. Simplified 20.1%

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

   4. Taylor expanded in A around -inf 1.0%

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

      1. *-commutative 1.0%

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

   6. Simplified 1.0%

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

   7. Taylor expanded in B around inf 4.6%

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

      1. *-commutative 4.6%

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

   9. Simplified 4.6%

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

4. Final simplification 12.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 3.2 \cdot 10^{+43}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(B \cdot B + -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot \left(2 \cdot A\right)\right)\right)}}{B \cdot B + -4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \]

Alternative 9: 22.7% accurate, 5.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;B \leq 1.1 \cdot 10^{+58}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (if (<= B 1.1e+58)
   (/
    (- (sqrt (* 2.0 (* (* 2.0 A) (* -4.0 (* A (* C F)))))))
    (- (* B B) (* (* A C) 4.0)))
   (* -2.0 (* (sqrt (* A F)) (/ 1.0 B)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (B <= 1.1e+58) {
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - ((A * C) * 4.0));
	} else {
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / 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) :: tmp
    if (b <= 1.1d+58) then
        tmp = -sqrt((2.0d0 * ((2.0d0 * a) * ((-4.0d0) * (a * (c * f)))))) / ((b * b) - ((a * c) * 4.0d0))
    else
        tmp = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / 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 tmp;
	if (B <= 1.1e+58) {
		tmp = -Math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - ((A * C) * 4.0));
	} else {
		tmp = -2.0 * (Math.sqrt((A * F)) * (1.0 / B));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if B <= 1.1e+58:
		tmp = -math.sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - ((A * C) * 4.0))
	else:
		tmp = -2.0 * (math.sqrt((A * F)) * (1.0 / B))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (B <= 1.1e+58)
		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(Float64(A * C) * 4.0)));
	else
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / B)));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	tmp = 0.0;
	if (B <= 1.1e+58)
		tmp = -sqrt((2.0 * ((2.0 * A) * (-4.0 * (A * (C * F)))))) / ((B * B) - ((A * C) * 4.0));
	else
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := If[LessEqual[B, 1.1e+58], 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[(N[(A * C), $MachinePrecision] * 4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if B < 1.1e58

   1. Initial program 18.1%

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

   3. Simplified 18.1%

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

   4. Taylor expanded in A around -inf 15.1%

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

      1. *-commutative 15.1%

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

   6. Simplified 15.1%

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

   7. Taylor expanded in B around 0 11.7%

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

      1. *-commutative 11.7%

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

   9. Simplified 11.7%

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

   if 1.1e58 < B

   1. Initial program 17.1%

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

   3. Simplified 17.1%

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

   4. Taylor expanded in A around -inf 0.9%

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

      1. *-commutative 0.9%

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

   6. Simplified 0.9%

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

   7. Taylor expanded in B around inf 4.8%

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

      1. *-commutative 4.8%

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

   9. Simplified 4.8%

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

4. Final simplification 10.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.1 \cdot 10^{+58}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(2 \cdot A\right) \cdot \left(-4 \cdot \left(A \cdot \left(C \cdot F\right)\right)\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \mathbf{else}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \end{array} \]

Alternative 10: 18.9% accurate, 5.1× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} \mathbf{if}\;A \leq -3.3 \cdot 10^{+153}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(-8 \cdot \left(\left(C \cdot F\right) \cdot \left(A \cdot A\right)\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \end{array} \end{array} \]

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 -3.3e+153)
   (* -2.0 (* (sqrt (* A F)) (/ 1.0 B)))
   (-
    (/
     (sqrt (* 2.0 (* -8.0 (* (* C F) (* A A)))))
     (- (* B B) (* (* A C) 4.0))))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double tmp;
	if (A <= -3.3e+153) {
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	} else {
		tmp = -(sqrt((2.0 * (-8.0 * ((C * F) * (A * A))))) / ((B * B) - ((A * C) * 4.0)));
	}
	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) :: tmp
    if (a <= (-3.3d+153)) then
        tmp = (-2.0d0) * (sqrt((a * f)) * (1.0d0 / b))
    else
        tmp = -(sqrt((2.0d0 * ((-8.0d0) * ((c * f) * (a * a))))) / ((b * b) - ((a * c) * 4.0d0)))
    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 <= -3.3e+153) {
		tmp = -2.0 * (Math.sqrt((A * F)) * (1.0 / B));
	} else {
		tmp = -(Math.sqrt((2.0 * (-8.0 * ((C * F) * (A * A))))) / ((B * B) - ((A * C) * 4.0)));
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	tmp = 0
	if A <= -3.3e+153:
		tmp = -2.0 * (math.sqrt((A * F)) * (1.0 / B))
	else:
		tmp = -(math.sqrt((2.0 * (-8.0 * ((C * F) * (A * A))))) / ((B * B) - ((A * C) * 4.0)))
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	tmp = 0.0
	if (A <= -3.3e+153)
		tmp = Float64(-2.0 * Float64(sqrt(Float64(A * F)) * Float64(1.0 / B)));
	else
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(-8.0 * Float64(Float64(C * F) * Float64(A * A))))) / Float64(Float64(B * B) - Float64(Float64(A * C) * 4.0))));
	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 <= -3.3e+153)
		tmp = -2.0 * (sqrt((A * F)) * (1.0 / B));
	else
		tmp = -(sqrt((2.0 * (-8.0 * ((C * F) * (A * A))))) / ((B * B) - ((A * C) * 4.0)));
	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_] := If[LessEqual[A, -3.3e+153], N[(-2.0 * N[(N[Sqrt[N[(A * F), $MachinePrecision]], $MachinePrecision] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], (-N[(N[Sqrt[N[(2.0 * N[(-8.0 * N[(N[(C * F), $MachinePrecision] * N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(N[(B * B), $MachinePrecision] - N[(N[(A * C), $MachinePrecision] * 4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision])]

Derivation

1. Split input into 2 regimes

2. if A < -3.29999999999999994e153

   1. Initial program 0.9%

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

   3. Simplified 0.9%

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

   4. Taylor expanded in A around -inf 10.9%

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

      1. *-commutative 10.9%

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

   6. Simplified 10.9%

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

   7. Taylor expanded in B around inf 5.0%

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

      1. *-commutative 5.0%

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

   9. Simplified 5.0%

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

   if -3.29999999999999994e153 < A

   1. Initial program 20.1%

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

   3. Simplified 20.1%

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

   4. Taylor expanded in A around -inf 12.6%

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

      1. *-commutative 12.6%

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

   6. Simplified 12.6%

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

   7. Taylor expanded in B around 0 9.9%

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

      1. unpow2 9.9%

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

      2. *-commutative 9.9%

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

   9. Simplified 9.9%

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

4. Final simplification 9.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;A \leq -3.3 \cdot 10^{+153}:\\ \;\;\;\;-2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right)\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(-8 \cdot \left(\left(C \cdot F\right) \cdot \left(A \cdot A\right)\right)\right)}}{B \cdot B - \left(A \cdot C\right) \cdot 4}\\ \end{array} \]

Alternative 11: 8.9% accurate, 5.8× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ -2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right) \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F) :precision binary64 (* -2.0 (* (sqrt (* A F)) (/ 1.0 B))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	return -2.0 * (sqrt((A * F)) * (1.0 / 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)) * (1.0d0 / 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)) * (1.0 / B));
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	return -2.0 * (math.sqrt((A * F)) * (1.0 / 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)) * Float64(1.0 / 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)) * (1.0 / 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] * N[(1.0 / B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 18.0%

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

3. Simplified 18.0%

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

4. Taylor expanded in A around -inf 12.4%

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

   1. *-commutative 12.4%

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

6. Simplified 12.4%

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

7. Taylor expanded in B around inf 2.4%

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

   1. *-commutative 2.4%

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

9. Simplified 2.4%

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

10. Final simplification 2.4%

    \[\leadsto -2 \cdot \left(\sqrt{A \cdot F} \cdot \frac{1}{B}\right) \]

Alternative 12: 8.9% accurate, 5.9× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 18.0%

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

3. Simplified 18.0%

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

4. Taylor expanded in A around -inf 12.4%

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

   1. *-commutative 12.4%

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

6. Simplified 12.4%

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

7. Taylor expanded in B around inf 2.4%

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

   1. associate-*r/ 2.4%

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

   2. *-rgt-identity 2.4%

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

   3. *-commutative 2.4%

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

9. Simplified 2.4%

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

10. Final simplification 2.4%

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

Reproduce

herbie shell --seed 2023230 
(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))))