ABCF->ab-angle a

Percentage Accurate: 18.6% → 50.3%
Time: 16.8s
Alternatives: 11
Speedup: 16.9×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t\_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t\_0} \end{array} \end{array} \]
(FPCore (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)))
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;
}
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
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;
}
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
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
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
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]]
\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}

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 11 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 18.6% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := {B}^{2} - \left(4 \cdot A\right) \cdot C\\ \frac{-\sqrt{\left(2 \cdot \left(t\_0 \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{t\_0} \end{array} \end{array} \]
(FPCore (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)))
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;
}
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
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;
}
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
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
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
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]]
\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}

Alternative 1: 50.3% accurate, 1.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := C \cdot \left(A \cdot 4\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(C \cdot 2\right) \cdot \left(\left(F \cdot \left({B\_m}^{2} - t\_0\right)\right) \cdot 2\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
B_m = (fabs.f64 B)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
(FPCore (A B_m C F)
 :precision binary64
 (let* ((t_0 (* C (* A 4.0))))
   (if (<= (pow B_m 2.0) 5e-85)
     (/
      (sqrt (* (* C 2.0) (* (* F (- (pow B_m 2.0) t_0)) 2.0)))
      (- t_0 (pow B_m 2.0)))
     (/ (sqrt (* F 2.0)) (- (sqrt B_m))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = C * (A * 4.0);
	double tmp;
	if (pow(B_m, 2.0) <= 5e-85) {
		tmp = sqrt(((C * 2.0) * ((F * (pow(B_m, 2.0) - t_0)) * 2.0))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	}
	return tmp;
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = c * (a * 4.0d0)
    if ((b_m ** 2.0d0) <= 5d-85) then
        tmp = sqrt(((c * 2.0d0) * ((f * ((b_m ** 2.0d0) - t_0)) * 2.0d0))) / (t_0 - (b_m ** 2.0d0))
    else
        tmp = sqrt((f * 2.0d0)) / -sqrt(b_m)
    end if
    code = tmp
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = C * (A * 4.0);
	double tmp;
	if (Math.pow(B_m, 2.0) <= 5e-85) {
		tmp = Math.sqrt(((C * 2.0) * ((F * (Math.pow(B_m, 2.0) - t_0)) * 2.0))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F * 2.0)) / -Math.sqrt(B_m);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = C * (A * 4.0)
	tmp = 0
	if math.pow(B_m, 2.0) <= 5e-85:
		tmp = math.sqrt(((C * 2.0) * ((F * (math.pow(B_m, 2.0) - t_0)) * 2.0))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F * 2.0)) / -math.sqrt(B_m)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(C * Float64(A * 4.0))
	tmp = 0.0
	if ((B_m ^ 2.0) <= 5e-85)
		tmp = Float64(sqrt(Float64(Float64(C * 2.0) * Float64(Float64(F * Float64((B_m ^ 2.0) - t_0)) * 2.0))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = C * (A * 4.0);
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 5e-85)
		tmp = sqrt(((C * 2.0) * ((F * ((B_m ^ 2.0) - t_0)) * 2.0))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(C * N[(A * 4.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-85], N[(N[Sqrt[N[(N[(C * 2.0), $MachinePrecision] * N[(N[(F * N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision]), $MachinePrecision] * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]
\begin{array}{l}
B_m = \left|B\right|
\\
[A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
\\
\begin{array}{l}
t_0 := C \cdot \left(A \cdot 4\right)\\
\mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\
\;\;\;\;\frac{\sqrt{\left(C \cdot 2\right) \cdot \left(\left(F \cdot \left({B\_m}^{2} - t\_0\right)\right) \cdot 2\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000002e-85

    1. Initial program 25.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in C around inf

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

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. lower-*.f6432.8

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites32.8%

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

    if 5.0000000000000002e-85 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 15.2%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Taylor expanded in B around inf

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

        \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
      2. *-commutativeN/A

        \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
      3. distribute-lft-neg-inN/A

        \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
      4. lower-*.f64N/A

        \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
      5. lower-neg.f64N/A

        \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
      6. lower-sqrt.f64N/A

        \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
      7. lower-sqrt.f64N/A

        \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
      8. lower-/.f6421.3

        \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
    5. Applied rewrites21.3%

      \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. Applied rewrites28.7%

        \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
      2. Step-by-step derivation
        1. Applied rewrites28.8%

          \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
      3. Recombined 2 regimes into one program.
      4. Final simplification30.5%

        \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(C \cdot 2\right) \cdot \left(\left(F \cdot \left({B}^{2} - C \cdot \left(A \cdot 4\right)\right)\right) \cdot 2\right)}}{C \cdot \left(A \cdot 4\right) - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
      5. Add Preprocessing

      Alternative 2: 50.2% accurate, 2.6× speedup?

      \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{-1}{t\_0} \cdot \sqrt{\left(t\_0 \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
      B_m = (fabs.f64 B)
      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
      (FPCore (A B_m C F)
       :precision binary64
       (let* ((t_0 (fma -4.0 (* C A) (* B_m B_m))))
         (if (<= (pow B_m 2.0) 5e-85)
           (* (/ -1.0 t_0) (sqrt (* (* t_0 (* F 2.0)) (* C 2.0))))
           (/ (sqrt (* F 2.0)) (- (sqrt B_m))))))
      B_m = fabs(B);
      assert(A < B_m && B_m < C && C < F);
      double code(double A, double B_m, double C, double F) {
      	double t_0 = fma(-4.0, (C * A), (B_m * B_m));
      	double tmp;
      	if (pow(B_m, 2.0) <= 5e-85) {
      		tmp = (-1.0 / t_0) * sqrt(((t_0 * (F * 2.0)) * (C * 2.0)));
      	} else {
      		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
      	}
      	return tmp;
      }
      
      B_m = abs(B)
      A, B_m, C, F = sort([A, B_m, C, F])
      function code(A, B_m, C, F)
      	t_0 = fma(-4.0, Float64(C * A), Float64(B_m * B_m))
      	tmp = 0.0
      	if ((B_m ^ 2.0) <= 5e-85)
      		tmp = Float64(Float64(-1.0 / t_0) * sqrt(Float64(Float64(t_0 * Float64(F * 2.0)) * Float64(C * 2.0))));
      	else
      		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
      	end
      	return tmp
      end
      
      B_m = N[Abs[B], $MachinePrecision]
      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
      code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(-4.0 * N[(C * A), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-85], N[(N[(-1.0 / t$95$0), $MachinePrecision] * N[Sqrt[N[(N[(t$95$0 * N[(F * 2.0), $MachinePrecision]), $MachinePrecision] * N[(C * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]
      
      \begin{array}{l}
      B_m = \left|B\right|
      \\
      [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
      \\
      \begin{array}{l}
      t_0 := \mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)\\
      \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\
      \;\;\;\;\frac{-1}{t\_0} \cdot \sqrt{\left(t\_0 \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}\\
      
      \mathbf{else}:\\
      \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 2 regimes
      2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000002e-85

        1. Initial program 25.2%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Add Preprocessing
        3. Applied rewrites25.2%

          \[\leadsto \color{blue}{\sqrt{\left(\left(C + A\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}} \]
        4. Taylor expanded in C around inf

          \[\leadsto \sqrt{\color{blue}{\left(C \cdot \left(2 + \left(-1 \cdot \frac{A}{C} + \frac{A}{C}\right)\right)\right)} \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
        5. Step-by-step derivation
          1. distribute-lft1-inN/A

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

            \[\leadsto \sqrt{\left(C \cdot \left(2 + \color{blue}{0} \cdot \frac{A}{C}\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
          3. mul0-lftN/A

            \[\leadsto \sqrt{\left(C \cdot \left(2 + \color{blue}{0}\right)\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
          4. metadata-evalN/A

            \[\leadsto \sqrt{\left(C \cdot \color{blue}{2}\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
          5. lower-*.f6432.7

            \[\leadsto \sqrt{\color{blue}{\left(C \cdot 2\right)} \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
        6. Applied rewrites32.7%

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

        if 5.0000000000000002e-85 < (pow.f64 B #s(literal 2 binary64))

        1. Initial program 15.2%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Add Preprocessing
        3. Taylor expanded in B around inf

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

            \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
          2. *-commutativeN/A

            \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
          3. distribute-lft-neg-inN/A

            \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
          4. lower-*.f64N/A

            \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
          5. lower-neg.f64N/A

            \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
          6. lower-sqrt.f64N/A

            \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
          7. lower-sqrt.f64N/A

            \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
          8. lower-/.f6421.3

            \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
        5. Applied rewrites21.3%

          \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
        6. Step-by-step derivation
          1. Applied rewrites28.7%

            \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
          2. Step-by-step derivation
            1. Applied rewrites28.8%

              \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
          3. Recombined 2 regimes into one program.
          4. Final simplification30.5%

            \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \cdot \sqrt{\left(\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
          5. Add Preprocessing

          Alternative 3: 50.3% accurate, 2.7× speedup?

          \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(-4 \cdot C, A, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(t\_0 \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
          B_m = (fabs.f64 B)
          NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
          (FPCore (A B_m C F)
           :precision binary64
           (let* ((t_0 (fma (* -4.0 C) A (* B_m B_m))))
             (if (<= (pow B_m 2.0) 5e-85)
               (/ (sqrt (* (* t_0 (* F 2.0)) (* C 2.0))) (- t_0))
               (/ (sqrt (* F 2.0)) (- (sqrt B_m))))))
          B_m = fabs(B);
          assert(A < B_m && B_m < C && C < F);
          double code(double A, double B_m, double C, double F) {
          	double t_0 = fma((-4.0 * C), A, (B_m * B_m));
          	double tmp;
          	if (pow(B_m, 2.0) <= 5e-85) {
          		tmp = sqrt(((t_0 * (F * 2.0)) * (C * 2.0))) / -t_0;
          	} else {
          		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
          	}
          	return tmp;
          }
          
          B_m = abs(B)
          A, B_m, C, F = sort([A, B_m, C, F])
          function code(A, B_m, C, F)
          	t_0 = fma(Float64(-4.0 * C), A, Float64(B_m * B_m))
          	tmp = 0.0
          	if ((B_m ^ 2.0) <= 5e-85)
          		tmp = Float64(sqrt(Float64(Float64(t_0 * Float64(F * 2.0)) * Float64(C * 2.0))) / Float64(-t_0));
          	else
          		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
          	end
          	return tmp
          end
          
          B_m = N[Abs[B], $MachinePrecision]
          NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
          code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(-4.0 * C), $MachinePrecision] * A + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-85], N[(N[Sqrt[N[(N[(t$95$0 * N[(F * 2.0), $MachinePrecision]), $MachinePrecision] * N[(C * 2.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]]
          
          \begin{array}{l}
          B_m = \left|B\right|
          \\
          [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
          \\
          \begin{array}{l}
          t_0 := \mathsf{fma}\left(-4 \cdot C, A, B\_m \cdot B\_m\right)\\
          \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-85}:\\
          \;\;\;\;\frac{\sqrt{\left(t\_0 \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}}{-t\_0}\\
          
          \mathbf{else}:\\
          \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\
          
          
          \end{array}
          \end{array}
          
          Derivation
          1. Split input into 2 regimes
          2. if (pow.f64 B #s(literal 2 binary64)) < 5.0000000000000002e-85

            1. Initial program 25.2%

              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. Add Preprocessing
            3. Taylor expanded in C around inf

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

                \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              2. lower-*.f6432.8

                \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            5. Applied rewrites32.8%

              \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            6. Step-by-step derivation
              1. lift-/.f64N/A

                \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C \cdot 2\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C}} \]
              2. frac-2negN/A

                \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C \cdot 2\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
              3. lift-neg.f64N/A

                \[\leadsto \frac{\mathsf{neg}\left(\color{blue}{\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C \cdot 2\right)}\right)\right)}\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)} \]
              4. remove-double-negN/A

                \[\leadsto \frac{\color{blue}{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(C \cdot 2\right)}}}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)} \]
            7. Applied rewrites32.7%

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

            if 5.0000000000000002e-85 < (pow.f64 B #s(literal 2 binary64))

            1. Initial program 15.2%

              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. Add Preprocessing
            3. Taylor expanded in B around inf

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

                \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
              2. *-commutativeN/A

                \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
              3. distribute-lft-neg-inN/A

                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
              4. lower-*.f64N/A

                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
              5. lower-neg.f64N/A

                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
              6. lower-sqrt.f64N/A

                \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
              7. lower-sqrt.f64N/A

                \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
              8. lower-/.f6421.3

                \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
            5. Applied rewrites21.3%

              \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
            6. Step-by-step derivation
              1. Applied rewrites28.7%

                \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
              2. Step-by-step derivation
                1. Applied rewrites28.8%

                  \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
              3. Recombined 2 regimes into one program.
              4. Final simplification30.5%

                \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-85}:\\ \;\;\;\;\frac{\sqrt{\left(\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot 2\right)}}{-\mathsf{fma}\left(-4 \cdot C, A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
              5. Add Preprocessing

              Alternative 4: 43.4% accurate, 6.9× speedup?

              \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B\_m \leq 1.85 \cdot 10^{-69}:\\ \;\;\;\;\sqrt{\left(\left(\left(C \cdot C\right) \cdot F\right) \cdot 16\right) \cdot \left(-A\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
              B_m = (fabs.f64 B)
              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
              (FPCore (A B_m C F)
               :precision binary64
               (if (<= B_m 1.85e-69)
                 (*
                  (sqrt (* (* (* (* C C) F) 16.0) (- A)))
                  (/ -1.0 (fma -4.0 (* C A) (* B_m B_m))))
                 (/ (sqrt (* F 2.0)) (- (sqrt B_m)))))
              B_m = fabs(B);
              assert(A < B_m && B_m < C && C < F);
              double code(double A, double B_m, double C, double F) {
              	double tmp;
              	if (B_m <= 1.85e-69) {
              		tmp = sqrt(((((C * C) * F) * 16.0) * -A)) * (-1.0 / fma(-4.0, (C * A), (B_m * B_m)));
              	} else {
              		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
              	}
              	return tmp;
              }
              
              B_m = abs(B)
              A, B_m, C, F = sort([A, B_m, C, F])
              function code(A, B_m, C, F)
              	tmp = 0.0
              	if (B_m <= 1.85e-69)
              		tmp = Float64(sqrt(Float64(Float64(Float64(Float64(C * C) * F) * 16.0) * Float64(-A))) * Float64(-1.0 / fma(-4.0, Float64(C * A), Float64(B_m * B_m))));
              	else
              		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
              	end
              	return tmp
              end
              
              B_m = N[Abs[B], $MachinePrecision]
              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
              code[A_, B$95$m_, C_, F_] := If[LessEqual[B$95$m, 1.85e-69], N[(N[Sqrt[N[(N[(N[(N[(C * C), $MachinePrecision] * F), $MachinePrecision] * 16.0), $MachinePrecision] * (-A)), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / N[(-4.0 * N[(C * A), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]
              
              \begin{array}{l}
              B_m = \left|B\right|
              \\
              [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
              \\
              \begin{array}{l}
              \mathbf{if}\;B\_m \leq 1.85 \cdot 10^{-69}:\\
              \;\;\;\;\sqrt{\left(\left(\left(C \cdot C\right) \cdot F\right) \cdot 16\right) \cdot \left(-A\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)}\\
              
              \mathbf{else}:\\
              \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\
              
              
              \end{array}
              \end{array}
              
              Derivation
              1. Split input into 2 regimes
              2. if B < 1.8500000000000001e-69

                1. Initial program 19.7%

                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                2. Add Preprocessing
                3. Applied rewrites19.7%

                  \[\leadsto \color{blue}{\sqrt{\left(\left(C + A\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}} \]
                4. Taylor expanded in A around -inf

                  \[\leadsto \sqrt{\color{blue}{-1 \cdot \left(A \cdot \left(-8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A} + 16 \cdot \left({C}^{2} \cdot F\right)\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                5. Step-by-step derivation
                  1. associate-*r*N/A

                    \[\leadsto \sqrt{\color{blue}{\left(-1 \cdot A\right) \cdot \left(-8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A} + 16 \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  2. lower-*.f64N/A

                    \[\leadsto \sqrt{\color{blue}{\left(-1 \cdot A\right) \cdot \left(-8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A} + 16 \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  3. mul-1-negN/A

                    \[\leadsto \sqrt{\color{blue}{\left(\mathsf{neg}\left(A\right)\right)} \cdot \left(-8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A} + 16 \cdot \left({C}^{2} \cdot F\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  4. lower-neg.f64N/A

                    \[\leadsto \sqrt{\color{blue}{\left(\mathsf{neg}\left(A\right)\right)} \cdot \left(-8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A} + 16 \cdot \left({C}^{2} \cdot F\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  5. +-commutativeN/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \color{blue}{\left(16 \cdot \left({C}^{2} \cdot F\right) + -8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  6. lower-fma.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \color{blue}{\mathsf{fma}\left(16, {C}^{2} \cdot F, -8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  7. lower-*.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \color{blue}{{C}^{2} \cdot F}, -8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  8. unpow2N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \color{blue}{\left(C \cdot C\right)} \cdot F, -8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  9. lower-*.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \color{blue}{\left(C \cdot C\right)} \cdot F, -8 \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  10. rem-square-sqrtN/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, \color{blue}{\left(\sqrt{-8} \cdot \sqrt{-8}\right)} \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  11. unpow2N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, \color{blue}{{\left(\sqrt{-8}\right)}^{2}} \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  12. lower-*.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, \color{blue}{{\left(\sqrt{-8}\right)}^{2} \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  13. unpow2N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, \color{blue}{\left(\sqrt{-8} \cdot \sqrt{-8}\right)} \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  14. rem-square-sqrtN/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, \color{blue}{-8} \cdot \frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  15. lower-/.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \color{blue}{\frac{{B}^{2} \cdot \left(C \cdot F\right)}{A}}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  16. associate-*r*N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \frac{\color{blue}{\left({B}^{2} \cdot C\right) \cdot F}}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  17. lower-*.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \frac{\color{blue}{\left({B}^{2} \cdot C\right) \cdot F}}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  18. lower-*.f64N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \frac{\color{blue}{\left({B}^{2} \cdot C\right)} \cdot F}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  19. unpow2N/A

                    \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \frac{\left(\color{blue}{\left(B \cdot B\right)} \cdot C\right) \cdot F}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                  20. lower-*.f6413.2

                    \[\leadsto \sqrt{\left(-A\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \frac{\left(\color{blue}{\left(B \cdot B\right)} \cdot C\right) \cdot F}{A}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                6. Applied rewrites13.2%

                  \[\leadsto \sqrt{\color{blue}{\left(-A\right) \cdot \mathsf{fma}\left(16, \left(C \cdot C\right) \cdot F, -8 \cdot \frac{\left(\left(B \cdot B\right) \cdot C\right) \cdot F}{A}\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                7. Taylor expanded in C around inf

                  \[\leadsto \sqrt{\left(\mathsf{neg}\left(A\right)\right) \cdot \left(16 \cdot \color{blue}{\left({C}^{2} \cdot F\right)}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                8. Step-by-step derivation
                  1. Applied rewrites13.8%

                    \[\leadsto \sqrt{\left(-A\right) \cdot \left(\left(\left(C \cdot C\right) \cdot F\right) \cdot \color{blue}{16}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

                  if 1.8500000000000001e-69 < B

                  1. Initial program 19.2%

                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                  2. Add Preprocessing
                  3. Taylor expanded in B around inf

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

                      \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                    2. *-commutativeN/A

                      \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                    3. distribute-lft-neg-inN/A

                      \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                    4. lower-*.f64N/A

                      \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                    5. lower-neg.f64N/A

                      \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                    6. lower-sqrt.f64N/A

                      \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                    7. lower-sqrt.f64N/A

                      \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                    8. lower-/.f6443.2

                      \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                  5. Applied rewrites43.2%

                    \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                  6. Step-by-step derivation
                    1. Applied rewrites59.2%

                      \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
                    2. Step-by-step derivation
                      1. Applied rewrites59.4%

                        \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
                    3. Recombined 2 regimes into one program.
                    4. Final simplification26.2%

                      \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.85 \cdot 10^{-69}:\\ \;\;\;\;\sqrt{\left(\left(\left(C \cdot C\right) \cdot F\right) \cdot 16\right) \cdot \left(-A\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
                    5. Add Preprocessing

                    Alternative 5: 43.4% accurate, 7.1× speedup?

                    \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;B\_m \leq 1.85 \cdot 10^{-69}:\\ \;\;\;\;\sqrt{\left(\left(\left(C \cdot C\right) \cdot F\right) \cdot A\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
                    B_m = (fabs.f64 B)
                    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                    (FPCore (A B_m C F)
                     :precision binary64
                     (if (<= B_m 1.85e-69)
                       (*
                        (sqrt (* (* (* (* C C) F) A) -16.0))
                        (/ -1.0 (fma -4.0 (* C A) (* B_m B_m))))
                       (/ (sqrt (* F 2.0)) (- (sqrt B_m)))))
                    B_m = fabs(B);
                    assert(A < B_m && B_m < C && C < F);
                    double code(double A, double B_m, double C, double F) {
                    	double tmp;
                    	if (B_m <= 1.85e-69) {
                    		tmp = sqrt(((((C * C) * F) * A) * -16.0)) * (-1.0 / fma(-4.0, (C * A), (B_m * B_m)));
                    	} else {
                    		tmp = sqrt((F * 2.0)) / -sqrt(B_m);
                    	}
                    	return tmp;
                    }
                    
                    B_m = abs(B)
                    A, B_m, C, F = sort([A, B_m, C, F])
                    function code(A, B_m, C, F)
                    	tmp = 0.0
                    	if (B_m <= 1.85e-69)
                    		tmp = Float64(sqrt(Float64(Float64(Float64(Float64(C * C) * F) * A) * -16.0)) * Float64(-1.0 / fma(-4.0, Float64(C * A), Float64(B_m * B_m))));
                    	else
                    		tmp = Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)));
                    	end
                    	return tmp
                    end
                    
                    B_m = N[Abs[B], $MachinePrecision]
                    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                    code[A_, B$95$m_, C_, F_] := If[LessEqual[B$95$m, 1.85e-69], N[(N[Sqrt[N[(N[(N[(N[(C * C), $MachinePrecision] * F), $MachinePrecision] * A), $MachinePrecision] * -16.0), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / N[(-4.0 * N[(C * A), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]]
                    
                    \begin{array}{l}
                    B_m = \left|B\right|
                    \\
                    [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                    \\
                    \begin{array}{l}
                    \mathbf{if}\;B\_m \leq 1.85 \cdot 10^{-69}:\\
                    \;\;\;\;\sqrt{\left(\left(\left(C \cdot C\right) \cdot F\right) \cdot A\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B\_m \cdot B\_m\right)}\\
                    
                    \mathbf{else}:\\
                    \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}\\
                    
                    
                    \end{array}
                    \end{array}
                    
                    Derivation
                    1. Split input into 2 regimes
                    2. if B < 1.8500000000000001e-69

                      1. Initial program 19.7%

                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                      2. Add Preprocessing
                      3. Applied rewrites19.7%

                        \[\leadsto \color{blue}{\sqrt{\left(\left(C + A\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}\right) \cdot \left(\left(2 \cdot F\right) \cdot \mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}} \]
                      4. Taylor expanded in C around inf

                        \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                      5. Step-by-step derivation
                        1. lower-*.f64N/A

                          \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                        2. lower-*.f64N/A

                          \[\leadsto \sqrt{-16 \cdot \color{blue}{\left(A \cdot \left({C}^{2} \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                        3. lower-*.f64N/A

                          \[\leadsto \sqrt{-16 \cdot \left(A \cdot \color{blue}{\left({C}^{2} \cdot F\right)}\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                        4. unpow2N/A

                          \[\leadsto \sqrt{-16 \cdot \left(A \cdot \left(\color{blue}{\left(C \cdot C\right)} \cdot F\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                        5. lower-*.f6413.7

                          \[\leadsto \sqrt{-16 \cdot \left(A \cdot \left(\color{blue}{\left(C \cdot C\right)} \cdot F\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]
                      6. Applied rewrites13.7%

                        \[\leadsto \sqrt{\color{blue}{-16 \cdot \left(A \cdot \left(\left(C \cdot C\right) \cdot F\right)\right)}} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)} \]

                      if 1.8500000000000001e-69 < B

                      1. Initial program 19.2%

                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                      2. Add Preprocessing
                      3. Taylor expanded in B around inf

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

                          \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                        2. *-commutativeN/A

                          \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                        3. distribute-lft-neg-inN/A

                          \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                        4. lower-*.f64N/A

                          \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                        5. lower-neg.f64N/A

                          \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                        6. lower-sqrt.f64N/A

                          \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                        7. lower-sqrt.f64N/A

                          \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                        8. lower-/.f6443.2

                          \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                      5. Applied rewrites43.2%

                        \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                      6. Step-by-step derivation
                        1. Applied rewrites59.2%

                          \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
                        2. Step-by-step derivation
                          1. Applied rewrites59.4%

                            \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
                        3. Recombined 2 regimes into one program.
                        4. Final simplification26.2%

                          \[\leadsto \begin{array}{l} \mathbf{if}\;B \leq 1.85 \cdot 10^{-69}:\\ \;\;\;\;\sqrt{\left(\left(\left(C \cdot C\right) \cdot F\right) \cdot A\right) \cdot -16} \cdot \frac{-1}{\mathsf{fma}\left(-4, C \cdot A, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{F \cdot 2}}{-\sqrt{B}}\\ \end{array} \]
                        5. Add Preprocessing

                        Alternative 6: 35.3% accurate, 12.6× speedup?

                        \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}} \end{array} \]
                        B_m = (fabs.f64 B)
                        NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                        (FPCore (A B_m C F) :precision binary64 (/ (sqrt (* F 2.0)) (- (sqrt B_m))))
                        B_m = fabs(B);
                        assert(A < B_m && B_m < C && C < F);
                        double code(double A, double B_m, double C, double F) {
                        	return sqrt((F * 2.0)) / -sqrt(B_m);
                        }
                        
                        B_m = abs(b)
                        NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                        real(8) function code(a, b_m, c, f)
                            real(8), intent (in) :: a
                            real(8), intent (in) :: b_m
                            real(8), intent (in) :: c
                            real(8), intent (in) :: f
                            code = sqrt((f * 2.0d0)) / -sqrt(b_m)
                        end function
                        
                        B_m = Math.abs(B);
                        assert A < B_m && B_m < C && C < F;
                        public static double code(double A, double B_m, double C, double F) {
                        	return Math.sqrt((F * 2.0)) / -Math.sqrt(B_m);
                        }
                        
                        B_m = math.fabs(B)
                        [A, B_m, C, F] = sort([A, B_m, C, F])
                        def code(A, B_m, C, F):
                        	return math.sqrt((F * 2.0)) / -math.sqrt(B_m)
                        
                        B_m = abs(B)
                        A, B_m, C, F = sort([A, B_m, C, F])
                        function code(A, B_m, C, F)
                        	return Float64(sqrt(Float64(F * 2.0)) / Float64(-sqrt(B_m)))
                        end
                        
                        B_m = abs(B);
                        A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                        function tmp = code(A, B_m, C, F)
                        	tmp = sqrt((F * 2.0)) / -sqrt(B_m);
                        end
                        
                        B_m = N[Abs[B], $MachinePrecision]
                        NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                        code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F * 2.0), $MachinePrecision]], $MachinePrecision] / (-N[Sqrt[B$95$m], $MachinePrecision])), $MachinePrecision]
                        
                        \begin{array}{l}
                        B_m = \left|B\right|
                        \\
                        [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                        \\
                        \frac{\sqrt{F \cdot 2}}{-\sqrt{B\_m}}
                        \end{array}
                        
                        Derivation
                        1. Initial program 19.6%

                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                        2. Add Preprocessing
                        3. Taylor expanded in B around inf

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

                            \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                          2. *-commutativeN/A

                            \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                          3. distribute-lft-neg-inN/A

                            \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                          4. lower-*.f64N/A

                            \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                          5. lower-neg.f64N/A

                            \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                          6. lower-sqrt.f64N/A

                            \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                          7. lower-sqrt.f64N/A

                            \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                          8. lower-/.f6412.8

                            \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                        5. Applied rewrites12.8%

                          \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                        6. Step-by-step derivation
                          1. Applied rewrites17.0%

                            \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
                          2. Step-by-step derivation
                            1. Applied rewrites17.0%

                              \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
                            2. Final simplification17.0%

                              \[\leadsto \frac{\sqrt{F \cdot 2}}{-\sqrt{B}} \]
                            3. Add Preprocessing

                            Alternative 7: 35.3% accurate, 12.6× speedup?

                            \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \left(-\sqrt{F}\right) \cdot \sqrt{\frac{2}{B\_m}} \end{array} \]
                            B_m = (fabs.f64 B)
                            NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                            (FPCore (A B_m C F) :precision binary64 (* (- (sqrt F)) (sqrt (/ 2.0 B_m))))
                            B_m = fabs(B);
                            assert(A < B_m && B_m < C && C < F);
                            double code(double A, double B_m, double C, double F) {
                            	return -sqrt(F) * sqrt((2.0 / B_m));
                            }
                            
                            B_m = abs(b)
                            NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                            real(8) function code(a, b_m, c, f)
                                real(8), intent (in) :: a
                                real(8), intent (in) :: b_m
                                real(8), intent (in) :: c
                                real(8), intent (in) :: f
                                code = -sqrt(f) * sqrt((2.0d0 / b_m))
                            end function
                            
                            B_m = Math.abs(B);
                            assert A < B_m && B_m < C && C < F;
                            public static double code(double A, double B_m, double C, double F) {
                            	return -Math.sqrt(F) * Math.sqrt((2.0 / B_m));
                            }
                            
                            B_m = math.fabs(B)
                            [A, B_m, C, F] = sort([A, B_m, C, F])
                            def code(A, B_m, C, F):
                            	return -math.sqrt(F) * math.sqrt((2.0 / B_m))
                            
                            B_m = abs(B)
                            A, B_m, C, F = sort([A, B_m, C, F])
                            function code(A, B_m, C, F)
                            	return Float64(Float64(-sqrt(F)) * sqrt(Float64(2.0 / B_m)))
                            end
                            
                            B_m = abs(B);
                            A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                            function tmp = code(A, B_m, C, F)
                            	tmp = -sqrt(F) * sqrt((2.0 / B_m));
                            end
                            
                            B_m = N[Abs[B], $MachinePrecision]
                            NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                            code[A_, B$95$m_, C_, F_] := N[((-N[Sqrt[F], $MachinePrecision]) * N[Sqrt[N[(2.0 / B$95$m), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
                            
                            \begin{array}{l}
                            B_m = \left|B\right|
                            \\
                            [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                            \\
                            \left(-\sqrt{F}\right) \cdot \sqrt{\frac{2}{B\_m}}
                            \end{array}
                            
                            Derivation
                            1. Initial program 19.6%

                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                            2. Add Preprocessing
                            3. Taylor expanded in B around inf

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

                                \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                              2. *-commutativeN/A

                                \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                              3. distribute-lft-neg-inN/A

                                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                              4. lower-*.f64N/A

                                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                              5. lower-neg.f64N/A

                                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                              6. lower-sqrt.f64N/A

                                \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                              7. lower-sqrt.f64N/A

                                \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                              8. lower-/.f6412.8

                                \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                            5. Applied rewrites12.8%

                              \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                            6. Step-by-step derivation
                              1. Applied rewrites12.9%

                                \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
                              2. Step-by-step derivation
                                1. Applied rewrites17.0%

                                  \[\leadsto -\sqrt{F} \cdot \sqrt{\frac{2}{B}} \]
                                2. Final simplification17.0%

                                  \[\leadsto \left(-\sqrt{F}\right) \cdot \sqrt{\frac{2}{B}} \]
                                3. Add Preprocessing

                                Alternative 8: 26.9% accurate, 12.9× speedup?

                                \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \frac{-2}{\sqrt{\frac{B\_m}{F} \cdot 2}} \end{array} \]
                                B_m = (fabs.f64 B)
                                NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                (FPCore (A B_m C F) :precision binary64 (/ -2.0 (sqrt (* (/ B_m F) 2.0))))
                                B_m = fabs(B);
                                assert(A < B_m && B_m < C && C < F);
                                double code(double A, double B_m, double C, double F) {
                                	return -2.0 / sqrt(((B_m / F) * 2.0));
                                }
                                
                                B_m = abs(b)
                                NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                real(8) function code(a, b_m, c, f)
                                    real(8), intent (in) :: a
                                    real(8), intent (in) :: b_m
                                    real(8), intent (in) :: c
                                    real(8), intent (in) :: f
                                    code = (-2.0d0) / sqrt(((b_m / f) * 2.0d0))
                                end function
                                
                                B_m = Math.abs(B);
                                assert A < B_m && B_m < C && C < F;
                                public static double code(double A, double B_m, double C, double F) {
                                	return -2.0 / Math.sqrt(((B_m / F) * 2.0));
                                }
                                
                                B_m = math.fabs(B)
                                [A, B_m, C, F] = sort([A, B_m, C, F])
                                def code(A, B_m, C, F):
                                	return -2.0 / math.sqrt(((B_m / F) * 2.0))
                                
                                B_m = abs(B)
                                A, B_m, C, F = sort([A, B_m, C, F])
                                function code(A, B_m, C, F)
                                	return Float64(-2.0 / sqrt(Float64(Float64(B_m / F) * 2.0)))
                                end
                                
                                B_m = abs(B);
                                A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                                function tmp = code(A, B_m, C, F)
                                	tmp = -2.0 / sqrt(((B_m / F) * 2.0));
                                end
                                
                                B_m = N[Abs[B], $MachinePrecision]
                                NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                code[A_, B$95$m_, C_, F_] := N[(-2.0 / N[Sqrt[N[(N[(B$95$m / F), $MachinePrecision] * 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
                                
                                \begin{array}{l}
                                B_m = \left|B\right|
                                \\
                                [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                                \\
                                \frac{-2}{\sqrt{\frac{B\_m}{F} \cdot 2}}
                                \end{array}
                                
                                Derivation
                                1. Initial program 19.6%

                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                2. Add Preprocessing
                                3. Taylor expanded in B around inf

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

                                    \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                                  2. *-commutativeN/A

                                    \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                                  3. distribute-lft-neg-inN/A

                                    \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                  4. lower-*.f64N/A

                                    \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                  5. lower-neg.f64N/A

                                    \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                                  6. lower-sqrt.f64N/A

                                    \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                                  7. lower-sqrt.f64N/A

                                    \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                                  8. lower-/.f6412.8

                                    \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                5. Applied rewrites12.8%

                                  \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                                6. Step-by-step derivation
                                  1. Applied rewrites17.0%

                                    \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
                                  2. Step-by-step derivation
                                    1. Applied rewrites17.0%

                                      \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
                                    2. Step-by-step derivation
                                      1. Applied rewrites12.8%

                                        \[\leadsto \frac{-2}{\color{blue}{\sqrt{\frac{B}{F} \cdot 2}}} \]
                                      2. Add Preprocessing

                                      Alternative 9: 27.3% accurate, 12.9× speedup?

                                      \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{\frac{\frac{F}{B\_m}}{2}} \cdot -2 \end{array} \]
                                      B_m = (fabs.f64 B)
                                      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                      (FPCore (A B_m C F) :precision binary64 (* (sqrt (/ (/ F B_m) 2.0)) -2.0))
                                      B_m = fabs(B);
                                      assert(A < B_m && B_m < C && C < F);
                                      double code(double A, double B_m, double C, double F) {
                                      	return sqrt(((F / B_m) / 2.0)) * -2.0;
                                      }
                                      
                                      B_m = abs(b)
                                      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                      real(8) function code(a, b_m, c, f)
                                          real(8), intent (in) :: a
                                          real(8), intent (in) :: b_m
                                          real(8), intent (in) :: c
                                          real(8), intent (in) :: f
                                          code = sqrt(((f / b_m) / 2.0d0)) * (-2.0d0)
                                      end function
                                      
                                      B_m = Math.abs(B);
                                      assert A < B_m && B_m < C && C < F;
                                      public static double code(double A, double B_m, double C, double F) {
                                      	return Math.sqrt(((F / B_m) / 2.0)) * -2.0;
                                      }
                                      
                                      B_m = math.fabs(B)
                                      [A, B_m, C, F] = sort([A, B_m, C, F])
                                      def code(A, B_m, C, F):
                                      	return math.sqrt(((F / B_m) / 2.0)) * -2.0
                                      
                                      B_m = abs(B)
                                      A, B_m, C, F = sort([A, B_m, C, F])
                                      function code(A, B_m, C, F)
                                      	return Float64(sqrt(Float64(Float64(F / B_m) / 2.0)) * -2.0)
                                      end
                                      
                                      B_m = abs(B);
                                      A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                                      function tmp = code(A, B_m, C, F)
                                      	tmp = sqrt(((F / B_m) / 2.0)) * -2.0;
                                      end
                                      
                                      B_m = N[Abs[B], $MachinePrecision]
                                      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                      code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(N[(F / B$95$m), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision] * -2.0), $MachinePrecision]
                                      
                                      \begin{array}{l}
                                      B_m = \left|B\right|
                                      \\
                                      [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                                      \\
                                      \sqrt{\frac{\frac{F}{B\_m}}{2}} \cdot -2
                                      \end{array}
                                      
                                      Derivation
                                      1. Initial program 19.6%

                                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                      2. Add Preprocessing
                                      3. Taylor expanded in B around inf

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

                                          \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                                        2. *-commutativeN/A

                                          \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                                        3. distribute-lft-neg-inN/A

                                          \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                        4. lower-*.f64N/A

                                          \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                        5. lower-neg.f64N/A

                                          \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                                        6. lower-sqrt.f64N/A

                                          \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                                        7. lower-sqrt.f64N/A

                                          \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                                        8. lower-/.f6412.8

                                          \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                      5. Applied rewrites12.8%

                                        \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                                      6. Step-by-step derivation
                                        1. Applied rewrites17.0%

                                          \[\leadsto \left(-\sqrt{2}\right) \cdot \frac{\sqrt{F}}{\color{blue}{\sqrt{B}}} \]
                                        2. Step-by-step derivation
                                          1. Applied rewrites17.0%

                                            \[\leadsto \frac{-\sqrt{F \cdot 2}}{\color{blue}{\sqrt{B}}} \]
                                          2. Step-by-step derivation
                                            1. Applied rewrites12.9%

                                              \[\leadsto -2 \cdot \color{blue}{\sqrt{\frac{\frac{F}{B}}{2}}} \]
                                            2. Final simplification12.9%

                                              \[\leadsto \sqrt{\frac{\frac{F}{B}}{2}} \cdot -2 \]
                                            3. Add Preprocessing

                                            Alternative 10: 27.3% accurate, 16.9× speedup?

                                            \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{\frac{F}{B\_m} \cdot 2} \end{array} \]
                                            B_m = (fabs.f64 B)
                                            NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                            (FPCore (A B_m C F) :precision binary64 (- (sqrt (* (/ F B_m) 2.0))))
                                            B_m = fabs(B);
                                            assert(A < B_m && B_m < C && C < F);
                                            double code(double A, double B_m, double C, double F) {
                                            	return -sqrt(((F / B_m) * 2.0));
                                            }
                                            
                                            B_m = abs(b)
                                            NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                            real(8) function code(a, b_m, c, f)
                                                real(8), intent (in) :: a
                                                real(8), intent (in) :: b_m
                                                real(8), intent (in) :: c
                                                real(8), intent (in) :: f
                                                code = -sqrt(((f / b_m) * 2.0d0))
                                            end function
                                            
                                            B_m = Math.abs(B);
                                            assert A < B_m && B_m < C && C < F;
                                            public static double code(double A, double B_m, double C, double F) {
                                            	return -Math.sqrt(((F / B_m) * 2.0));
                                            }
                                            
                                            B_m = math.fabs(B)
                                            [A, B_m, C, F] = sort([A, B_m, C, F])
                                            def code(A, B_m, C, F):
                                            	return -math.sqrt(((F / B_m) * 2.0))
                                            
                                            B_m = abs(B)
                                            A, B_m, C, F = sort([A, B_m, C, F])
                                            function code(A, B_m, C, F)
                                            	return Float64(-sqrt(Float64(Float64(F / B_m) * 2.0)))
                                            end
                                            
                                            B_m = abs(B);
                                            A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                                            function tmp = code(A, B_m, C, F)
                                            	tmp = -sqrt(((F / B_m) * 2.0));
                                            end
                                            
                                            B_m = N[Abs[B], $MachinePrecision]
                                            NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                            code[A_, B$95$m_, C_, F_] := (-N[Sqrt[N[(N[(F / B$95$m), $MachinePrecision] * 2.0), $MachinePrecision]], $MachinePrecision])
                                            
                                            \begin{array}{l}
                                            B_m = \left|B\right|
                                            \\
                                            [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                                            \\
                                            -\sqrt{\frac{F}{B\_m} \cdot 2}
                                            \end{array}
                                            
                                            Derivation
                                            1. Initial program 19.6%

                                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                            2. Add Preprocessing
                                            3. Taylor expanded in B around inf

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

                                                \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                                              2. *-commutativeN/A

                                                \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                                              3. distribute-lft-neg-inN/A

                                                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                              4. lower-*.f64N/A

                                                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                              5. lower-neg.f64N/A

                                                \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                                              6. lower-sqrt.f64N/A

                                                \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                                              7. lower-sqrt.f64N/A

                                                \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                                              8. lower-/.f6412.8

                                                \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                            5. Applied rewrites12.8%

                                              \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                                            6. Step-by-step derivation
                                              1. Applied rewrites12.9%

                                                \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
                                              2. Add Preprocessing

                                              Alternative 11: 27.3% accurate, 16.9× speedup?

                                              \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{\frac{2}{B\_m} \cdot F} \end{array} \]
                                              B_m = (fabs.f64 B)
                                              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                              (FPCore (A B_m C F) :precision binary64 (- (sqrt (* (/ 2.0 B_m) F))))
                                              B_m = fabs(B);
                                              assert(A < B_m && B_m < C && C < F);
                                              double code(double A, double B_m, double C, double F) {
                                              	return -sqrt(((2.0 / B_m) * F));
                                              }
                                              
                                              B_m = abs(b)
                                              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                              real(8) function code(a, b_m, c, f)
                                                  real(8), intent (in) :: a
                                                  real(8), intent (in) :: b_m
                                                  real(8), intent (in) :: c
                                                  real(8), intent (in) :: f
                                                  code = -sqrt(((2.0d0 / b_m) * f))
                                              end function
                                              
                                              B_m = Math.abs(B);
                                              assert A < B_m && B_m < C && C < F;
                                              public static double code(double A, double B_m, double C, double F) {
                                              	return -Math.sqrt(((2.0 / B_m) * F));
                                              }
                                              
                                              B_m = math.fabs(B)
                                              [A, B_m, C, F] = sort([A, B_m, C, F])
                                              def code(A, B_m, C, F):
                                              	return -math.sqrt(((2.0 / B_m) * F))
                                              
                                              B_m = abs(B)
                                              A, B_m, C, F = sort([A, B_m, C, F])
                                              function code(A, B_m, C, F)
                                              	return Float64(-sqrt(Float64(Float64(2.0 / B_m) * F)))
                                              end
                                              
                                              B_m = abs(B);
                                              A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                                              function tmp = code(A, B_m, C, F)
                                              	tmp = -sqrt(((2.0 / B_m) * F));
                                              end
                                              
                                              B_m = N[Abs[B], $MachinePrecision]
                                              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                                              code[A_, B$95$m_, C_, F_] := (-N[Sqrt[N[(N[(2.0 / B$95$m), $MachinePrecision] * F), $MachinePrecision]], $MachinePrecision])
                                              
                                              \begin{array}{l}
                                              B_m = \left|B\right|
                                              \\
                                              [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                                              \\
                                              -\sqrt{\frac{2}{B\_m} \cdot F}
                                              \end{array}
                                              
                                              Derivation
                                              1. Initial program 19.6%

                                                \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                              2. Add Preprocessing
                                              3. Taylor expanded in B around inf

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

                                                  \[\leadsto \color{blue}{\mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \sqrt{2}\right)} \]
                                                2. *-commutativeN/A

                                                  \[\leadsto \mathsf{neg}\left(\color{blue}{\sqrt{2} \cdot \sqrt{\frac{F}{B}}}\right) \]
                                                3. distribute-lft-neg-inN/A

                                                  \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                                4. lower-*.f64N/A

                                                  \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \sqrt{\frac{F}{B}}} \]
                                                5. lower-neg.f64N/A

                                                  \[\leadsto \color{blue}{\left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \cdot \sqrt{\frac{F}{B}} \]
                                                6. lower-sqrt.f64N/A

                                                  \[\leadsto \left(\mathsf{neg}\left(\color{blue}{\sqrt{2}}\right)\right) \cdot \sqrt{\frac{F}{B}} \]
                                                7. lower-sqrt.f64N/A

                                                  \[\leadsto \left(\mathsf{neg}\left(\sqrt{2}\right)\right) \cdot \color{blue}{\sqrt{\frac{F}{B}}} \]
                                                8. lower-/.f6412.8

                                                  \[\leadsto \left(-\sqrt{2}\right) \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                              5. Applied rewrites12.8%

                                                \[\leadsto \color{blue}{\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F}{B}}} \]
                                              6. Step-by-step derivation
                                                1. Applied rewrites12.9%

                                                  \[\leadsto \color{blue}{-\sqrt{\frac{F}{B} \cdot 2}} \]
                                                2. Step-by-step derivation
                                                  1. Applied rewrites12.9%

                                                    \[\leadsto -\sqrt{F \cdot \frac{2}{B}} \]
                                                  2. Final simplification12.9%

                                                    \[\leadsto -\sqrt{\frac{2}{B} \cdot F} \]
                                                  3. Add Preprocessing

                                                  Reproduce

                                                  ?
                                                  herbie shell --seed 2024235 
                                                  (FPCore (A B C F)
                                                    :name "ABCF->ab-angle a"
                                                    :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))))