ABCF->ab-angle b

Percentage Accurate: 18.0% → 45.8%

Time: 15.4s

Alternatives: 5

Speedup: 5.0×

Specification

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 18.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 45.8% accurate, 2.0× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ \mathbf{if}\;B \leq 7.5 \cdot 10^{+27}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(t_0 \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))))
   (if (<= B 7.5e+27)
     (/ (- (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 <= 7.5e+27) {
		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 <= 7.5e+27) {
		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 <= 7.5e+27:
		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 <= 7.5e+27)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(Float64(t_0 * F) * Float64(2.0 * A))))) / t_0);
	else
		tmp = Float64(sqrt(Float64(F * Float64(A - hypot(A, B)))) * Float64(Float64(-sqrt(2.0)) / B));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	tmp = 0.0;
	if (B <= 7.5e+27)
		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, 7.5e+27], N[((-N[Sqrt[N[(2.0 * N[(N[(t$95$0 * F), $MachinePrecision] * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[(N[Sqrt[N[(F * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[((-N[Sqrt[2.0], $MachinePrecision]) / B), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if B < 7.5000000000000002e27

   1. Initial program 17.2%

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

   3. Simplified 17.2%

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

   4. Taylor expanded in A around -inf 19.4%

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

   if 7.5000000000000002e27 < B

   1. Initial program 11.6%

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

   3. Simplified 11.6%

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

   4. Taylor expanded in C around 0 26.3%

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

      1. mul-1-neg 26.3%

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

      2. +-commutative 26.3%

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

      3. unpow2 26.3%

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

      4. unpow2 26.3%

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

      5. hypot-def 63.3%

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

   6. Simplified 63.3%

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

4. Final simplification 30.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 7.5 \cdot 10^{+27}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{F \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)} \cdot \frac{-\sqrt{2}}{B}\\ \end{array} \]

Alternative 2: 35.2% accurate, 2.7× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;C \leq 3.8 \cdot 10^{+15}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* t_0 F)))
   (if (<= C 3.8e+15)
     (/ (- (sqrt (* 2.0 (* t_1 (- A (hypot A B)))))) t_0)
     (/
      (-
       (sqrt
        (*
         2.0
         (* t_1 (+ A (+ A (* -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C))))))))
      t_0))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (C <= 3.8e+15) {
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
	} else {
		tmp = -sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	}
	return tmp;
}

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (C <= 3.8e+15) {
		tmp = -Math.sqrt((2.0 * (t_1 * (A - Math.hypot(A, B))))) / t_0;
	} else {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	t_1 = t_0 * F
	tmp = 0
	if C <= 3.8e+15:
		tmp = -math.sqrt((2.0 * (t_1 * (A - math.hypot(A, B))))) / t_0
	else:
		tmp = -math.sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_1 = Float64(t_0 * F)
	tmp = 0.0
	if (C <= 3.8e+15)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A - hypot(A, B)))))) / t_0);
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + Float64(A + Float64(-0.5 * Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C)))))))) / t_0);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	t_1 = t_0 * F;
	tmp = 0.0;
	if (C <= 3.8e+15)
		tmp = -sqrt((2.0 * (t_1 * (A - hypot(A, B))))) / t_0;
	else
		tmp = -sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[C, 3.8e+15], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A - N[Sqrt[A ^ 2 + B ^ 2], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if C < 3.8e15

   1. Initial program 21.1%

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

   3. Simplified 21.1%

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

   4. Taylor expanded in C around 0 18.2%

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

      1. +-commutative 18.2%

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

      2. unpow2 18.2%

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

      3. unpow2 18.2%

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

      4. hypot-def 21.2%

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

   6. Simplified 21.2%

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

   if 3.8e15 < C

   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 C around inf 27.5%

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

      1. associate--l+ 27.5%

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

      2. unpow2 27.5%

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

      3. unpow2 27.5%

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

      4. unpow2 27.5%

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

      5. mul-1-neg 27.5%

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

      6. mul-1-neg 27.5%

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

      7. sqr-neg 27.5%

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

      8. mul-1-neg 27.5%

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

   6. Simplified 27.5%

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

4. Final simplification 22.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq 3.8 \cdot 10^{+15}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A - \mathsf{hypot}\left(A, B\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

Alternative 3: 28.4% accurate, 4.3× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;C \leq -4.8 \cdot 10^{-294}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{elif}\;C \leq 1.3 \cdot 10^{-15}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* t_0 F)))
   (if (<= C -4.8e-294)
     (/ (- (sqrt (* 2.0 (* t_1 (* 2.0 A))))) t_0)
     (if (<= C 1.3e-15)
       (- (/ (sqrt (* 2.0 (* t_1 (- (+ A C) B)))) t_0))
       (/
        (-
         (sqrt
          (*
           2.0
           (*
            t_1
            (+ A (+ A (* -0.5 (/ (+ (* B B) (- (* A A) (* A A))) C))))))))
        t_0)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (C <= -4.8e-294) {
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	} else if (C <= 1.3e-15) {
		tmp = -(sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0);
	} else {
		tmp = -sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (a * c))
    t_1 = t_0 * f
    if (c <= (-4.8d-294)) then
        tmp = -sqrt((2.0d0 * (t_1 * (2.0d0 * a)))) / t_0
    else if (c <= 1.3d-15) then
        tmp = -(sqrt((2.0d0 * (t_1 * ((a + c) - b)))) / t_0)
    else
        tmp = -sqrt((2.0d0 * (t_1 * (a + (a + ((-0.5d0) * (((b * b) + ((a * a) - (a * a))) / c))))))) / t_0
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (C <= -4.8e-294) {
		tmp = -Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	} else if (C <= 1.3e-15) {
		tmp = -(Math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0);
	} else {
		tmp = -Math.sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	t_1 = t_0 * F
	tmp = 0
	if C <= -4.8e-294:
		tmp = -math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0
	elif C <= 1.3e-15:
		tmp = -(math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0)
	else:
		tmp = -math.sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_1 = Float64(t_0 * F)
	tmp = 0.0
	if (C <= -4.8e-294)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A))))) / t_0);
	elseif (C <= 1.3e-15)
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(Float64(A + C) - B)))) / t_0));
	else
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(A + Float64(A + Float64(-0.5 * Float64(Float64(Float64(B * B) + Float64(Float64(A * A) - Float64(A * A))) / C)))))))) / t_0);
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	t_1 = t_0 * F;
	tmp = 0.0;
	if (C <= -4.8e-294)
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	elseif (C <= 1.3e-15)
		tmp = -(sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0);
	else
		tmp = -sqrt((2.0 * (t_1 * (A + (A + (-0.5 * (((B * B) + ((A * A) - (A * A))) / C))))))) / t_0;
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[C, -4.8e-294], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], If[LessEqual[C, 1.3e-15], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(N[(A + C), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision]), N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(A + N[(A + N[(-0.5 * N[(N[(N[(B * B), $MachinePrecision] + N[(N[(A * A), $MachinePrecision] - N[(A * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision]]]]]

Derivation

1. Split input into 3 regimes

2. if C < -4.79999999999999994e-294

   1. Initial program 20.2%

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

   3. Simplified 20.2%

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

   4. Taylor expanded in A around -inf 11.3%

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

   if -4.79999999999999994e-294 < C < 1.30000000000000002e-15

   1. Initial program 23.4%

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

   3. Simplified 23.4%

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

   4. Taylor expanded in B around inf 12.5%

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

   if 1.30000000000000002e-15 < C

   1. Initial program 3.6%

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

   3. Simplified 3.6%

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

   4. Taylor expanded in C around inf 27.3%

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

      1. associate--l+ 27.3%

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

      2. unpow2 27.3%

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

      3. unpow2 27.3%

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

      4. unpow2 27.3%

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

      5. mul-1-neg 27.3%

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

      6. mul-1-neg 27.3%

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

      7. sqr-neg 27.3%

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

      8. mul-1-neg 27.3%

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

   6. Simplified 27.3%

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

4. Final simplification 16.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;C \leq -4.8 \cdot 10^{-294}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{elif}\;C \leq 1.3 \cdot 10^{-15}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(A + \left(A + -0.5 \cdot \frac{B \cdot B + \left(A \cdot A - A \cdot A\right)}{C}\right)\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

Alternative 4: 28.0% accurate, 4.8× speedup

Math

\[\begin{array}{l} B = |B|\\ [A, C] = \mathsf{sort}([A, C])\\ \\ \begin{array}{l} t_0 := B \cdot B - 4 \cdot \left(A \cdot C\right)\\ t_1 := t_0 \cdot F\\ \mathbf{if}\;B \leq 1.4 \cdot 10^{+28}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(t_1 \cdot \left(2 \cdot A\right)\right)}}{t_0}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(t_1 \cdot \left(\left(A + C\right) - B\right)\right)}}{t_0}\\ \end{array} \end{array} \]

FPCore

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
(FPCore (A B C F)
 :precision binary64
 (let* ((t_0 (- (* B B) (* 4.0 (* A C)))) (t_1 (* t_0 F)))
   (if (<= B 1.4e+28)
     (/ (- (sqrt (* 2.0 (* t_1 (* 2.0 A))))) t_0)
     (- (/ (sqrt (* 2.0 (* t_1 (- (+ A C) B)))) t_0)))))

C

B = abs(B);
assert(A < C);
double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (B <= 1.4e+28) {
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	} else {
		tmp = -(sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0);
	}
	return tmp;
}

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: t_1
    real(8) :: tmp
    t_0 = (b * b) - (4.0d0 * (a * c))
    t_1 = t_0 * f
    if (b <= 1.4d+28) then
        tmp = -sqrt((2.0d0 * (t_1 * (2.0d0 * a)))) / t_0
    else
        tmp = -(sqrt((2.0d0 * (t_1 * ((a + c) - b)))) / t_0)
    end if
    code = tmp
end function

Java

B = Math.abs(B);
assert A < C;
public static double code(double A, double B, double C, double F) {
	double t_0 = (B * B) - (4.0 * (A * C));
	double t_1 = t_0 * F;
	double tmp;
	if (B <= 1.4e+28) {
		tmp = -Math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	} else {
		tmp = -(Math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0);
	}
	return tmp;
}

Python

B = abs(B)
[A, C] = sort([A, C])
def code(A, B, C, F):
	t_0 = (B * B) - (4.0 * (A * C))
	t_1 = t_0 * F
	tmp = 0
	if B <= 1.4e+28:
		tmp = -math.sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0
	else:
		tmp = -(math.sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0)
	return tmp

Julia

B = abs(B)
A, C = sort([A, C])
function code(A, B, C, F)
	t_0 = Float64(Float64(B * B) - Float64(4.0 * Float64(A * C)))
	t_1 = Float64(t_0 * F)
	tmp = 0.0
	if (B <= 1.4e+28)
		tmp = Float64(Float64(-sqrt(Float64(2.0 * Float64(t_1 * Float64(2.0 * A))))) / t_0);
	else
		tmp = Float64(-Float64(sqrt(Float64(2.0 * Float64(t_1 * Float64(Float64(A + C) - B)))) / t_0));
	end
	return tmp
end

MATLAB

B = abs(B)
A, C = num2cell(sort([A, C])){:}
function tmp_2 = code(A, B, C, F)
	t_0 = (B * B) - (4.0 * (A * C));
	t_1 = t_0 * F;
	tmp = 0.0;
	if (B <= 1.4e+28)
		tmp = -sqrt((2.0 * (t_1 * (2.0 * A)))) / t_0;
	else
		tmp = -(sqrt((2.0 * (t_1 * ((A + C) - B)))) / t_0);
	end
	tmp_2 = tmp;
end

Wolfram

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
code[A_, B_, C_, F_] := Block[{t$95$0 = N[(N[(B * B), $MachinePrecision] - N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(t$95$0 * F), $MachinePrecision]}, If[LessEqual[B, 1.4e+28], N[((-N[Sqrt[N[(2.0 * N[(t$95$1 * N[(2.0 * A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]) / t$95$0), $MachinePrecision], (-N[(N[Sqrt[N[(2.0 * N[(t$95$1 * N[(N[(A + C), $MachinePrecision] - B), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$0), $MachinePrecision])]]]

Derivation

1. Split input into 2 regimes

2. if B < 1.4000000000000001e28

   1. Initial program 17.2%

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

   3. Simplified 17.2%

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

   4. Taylor expanded in A around -inf 19.4%

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

   if 1.4000000000000001e28 < B

   1. Initial program 11.6%

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

   3. Simplified 11.6%

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

   4. Taylor expanded in B around inf 10.2%

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

4. Final simplification 17.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.4 \cdot 10^{+28}:\\ \;\;\;\;\frac{-\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(2 \cdot A\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;-\frac{\sqrt{2 \cdot \left(\left(\left(B \cdot B - 4 \cdot \left(A \cdot C\right)\right) \cdot F\right) \cdot \left(\left(A + C\right) - B\right)\right)}}{B \cdot B - 4 \cdot \left(A \cdot C\right)}\\ \end{array} \]

Alternative 5: 26.0% accurate, 5.0× speedup

Math

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

FPCore

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

C

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

Fortran

NOTE: B should be positive before calling this function
NOTE: A and C should be sorted in increasing order before calling this function.
real(8) function code(a, b, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    t_0 = (b * b) - (4.0d0 * (a * c))
    code = -sqrt((2.0d0 * ((t_0 * f) * (2.0d0 * a)))) / t_0
end function

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 15.8%

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

3. Simplified 15.8%

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

4. Taylor expanded in A around -inf 15.5%

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

5. Final simplification 15.5%

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

Reproduce

herbie shell --seed 2023238 
(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))))