ABCF->ab-angle a

Percentage Accurate: 18.8% → 57.1%
Time: 19.2s
Alternatives: 12
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 12 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: 18.8% 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: 57.1% accurate, 0.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\ t_2 := \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\\ t_3 := \left(4 \cdot A\right) \cdot C\\ t_4 := t\_3 - {B\_m}^{2}\\ t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4}\\ \mathbf{if}\;t\_5 \leq -\infty:\\ \;\;\;\;\frac{\sqrt{2 \cdot t\_0} \cdot \sqrt{F \cdot t\_2}}{t\_4}\\ \mathbf{elif}\;t\_5 \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\ \mathbf{elif}\;t\_5 \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot t\_2\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
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 C (* A -4.0) (* B_m B_m)))
        (t_1 (fma B_m B_m (* -4.0 (* A C))))
        (t_2 (fma 2.0 C (/ (* (* B_m B_m) -0.5) A)))
        (t_3 (* (* 4.0 A) C))
        (t_4 (- t_3 (pow B_m 2.0)))
        (t_5
         (/
          (sqrt
           (*
            (* 2.0 (* (- (pow B_m 2.0) t_3) F))
            (+ (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
          t_4)))
   (if (<= t_5 (- INFINITY))
     (/ (* (sqrt (* 2.0 t_0)) (sqrt (* F t_2))) t_4)
     (if (<= t_5 -2e-194)
       (*
        (/ (sqrt (* t_1 (* 2.0 F))) -1.0)
        (/ (sqrt (+ (+ A C) (sqrt (fma (- A C) (- A C) (* B_m B_m))))) t_1))
       (if (<= t_5 INFINITY)
         (/ (sqrt (* (* F t_0) (* 2.0 t_2))) (- t_0))
         (/ (sqrt (* 2.0 F)) (- (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(C, (A * -4.0), (B_m * B_m));
	double t_1 = fma(B_m, B_m, (-4.0 * (A * C)));
	double t_2 = fma(2.0, C, (((B_m * B_m) * -0.5) / A));
	double t_3 = (4.0 * A) * C;
	double t_4 = t_3 - pow(B_m, 2.0);
	double t_5 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_3) * F)) * ((A + C) + sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / t_4;
	double tmp;
	if (t_5 <= -((double) INFINITY)) {
		tmp = (sqrt((2.0 * t_0)) * sqrt((F * t_2))) / t_4;
	} else if (t_5 <= -2e-194) {
		tmp = (sqrt((t_1 * (2.0 * F))) / -1.0) * (sqrt(((A + C) + sqrt(fma((A - C), (A - C), (B_m * B_m))))) / t_1);
	} else if (t_5 <= ((double) INFINITY)) {
		tmp = sqrt(((F * t_0) * (2.0 * t_2))) / -t_0;
	} else {
		tmp = sqrt((2.0 * F)) / -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(C, Float64(A * -4.0), Float64(B_m * B_m))
	t_1 = fma(B_m, B_m, Float64(-4.0 * Float64(A * C)))
	t_2 = fma(2.0, C, Float64(Float64(Float64(B_m * B_m) * -0.5) / A))
	t_3 = Float64(Float64(4.0 * A) * C)
	t_4 = Float64(t_3 - (B_m ^ 2.0))
	t_5 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_3) * F)) * Float64(Float64(A + C) + sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / t_4)
	tmp = 0.0
	if (t_5 <= Float64(-Inf))
		tmp = Float64(Float64(sqrt(Float64(2.0 * t_0)) * sqrt(Float64(F * t_2))) / t_4);
	elseif (t_5 <= -2e-194)
		tmp = Float64(Float64(sqrt(Float64(t_1 * Float64(2.0 * F))) / -1.0) * Float64(sqrt(Float64(Float64(A + C) + sqrt(fma(Float64(A - C), Float64(A - C), Float64(B_m * B_m))))) / t_1));
	elseif (t_5 <= Inf)
		tmp = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * t_2))) / Float64(-t_0));
	else
		tmp = Float64(sqrt(Float64(2.0 * F)) / 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[(C * N[(A * -4.0), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(2.0 * C + N[(N[(N[(B$95$m * B$95$m), $MachinePrecision] * -0.5), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$4 = N[(t$95$3 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$5 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$3), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$4), $MachinePrecision]}, If[LessEqual[t$95$5, (-Infinity)], N[(N[(N[Sqrt[N[(2.0 * t$95$0), $MachinePrecision]], $MachinePrecision] * N[Sqrt[N[(F * t$95$2), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / t$95$4), $MachinePrecision], If[LessEqual[t$95$5, -2e-194], N[(N[(N[Sqrt[N[(t$95$1 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / -1.0), $MachinePrecision] * N[(N[Sqrt[N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[(A - C), $MachinePrecision] * N[(A - C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$5, Infinity], N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * t$95$2), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], N[(N[Sqrt[N[(2.0 * F), $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(C, A \cdot -4, B\_m \cdot B\_m\right)\\
t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\
t_2 := \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\\
t_3 := \left(4 \cdot A\right) \cdot C\\
t_4 := t\_3 - {B\_m}^{2}\\
t_5 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_4}\\
\mathbf{if}\;t\_5 \leq -\infty:\\
\;\;\;\;\frac{\sqrt{2 \cdot t\_0} \cdot \sqrt{F \cdot t\_2}}{t\_4}\\

\mathbf{elif}\;t\_5 \leq -2 \cdot 10^{-194}:\\
\;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\

\mathbf{elif}\;t\_5 \leq \infty:\\
\;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot t\_2\right)}}{-t\_0}\\

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


\end{array}
\end{array}
Derivation
  1. Split input into 4 regimes
  2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -inf.0

    1. Initial program 2.9%

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

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

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

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      7. lower-*.f6425.9

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites25.9%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Step-by-step derivation
      1. lift-sqrt.f64N/A

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

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

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\color{blue}{\sqrt{2 \cdot \left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)} \cdot \sqrt{F \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    7. Applied rewrites34.0%

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

    if -inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -2.00000000000000004e-194

    1. Initial program 97.6%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. Add Preprocessing
    3. Applied rewrites99.3%

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

    if -2.00000000000000004e-194 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < +inf.0

    1. Initial program 14.2%

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

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

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

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      7. lower-*.f6428.2

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites28.2%

      \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    6. Step-by-step derivation
      1. lift-/.f64N/A

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

        \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
    7. Applied rewrites28.2%

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

    if +inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)))

    1. Initial program 0.0%

      \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    2. 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. lower-neg.f64N/A

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

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

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

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

        \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
      7. lower-/.f6416.4

        \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
    5. Applied rewrites16.4%

      \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
    6. Step-by-step derivation
      1. Applied rewrites25.2%

        \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
    7. Recombined 4 regimes into one program.
    8. Final simplification37.7%

      \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\frac{\sqrt{2 \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)} \cdot \sqrt{F \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}}}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B}}\\ \end{array} \]
    9. Add Preprocessing

    Alternative 2: 54.2% accurate, 0.3× speedup?

    \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\ t_2 := \frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\ t_3 := \left(4 \cdot A\right) \cdot C\\ t_4 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_3 - {B\_m}^{2}}\\ \mathbf{if}\;t\_4 \leq -\infty:\\ \;\;\;\;t\_2\\ \mathbf{elif}\;t\_4 \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\ \mathbf{elif}\;t\_4 \leq \infty:\\ \;\;\;\;t\_2\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
    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 C (* A -4.0) (* B_m B_m)))
            (t_1 (fma B_m B_m (* -4.0 (* A C))))
            (t_2
             (/
              (sqrt (* (* F t_0) (* 2.0 (fma 2.0 C (/ (* (* B_m B_m) -0.5) A)))))
              (- t_0)))
            (t_3 (* (* 4.0 A) C))
            (t_4
             (/
              (sqrt
               (*
                (* 2.0 (* (- (pow B_m 2.0) t_3) F))
                (+ (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
              (- t_3 (pow B_m 2.0)))))
       (if (<= t_4 (- INFINITY))
         t_2
         (if (<= t_4 -2e-194)
           (*
            (/ (sqrt (* t_1 (* 2.0 F))) -1.0)
            (/ (sqrt (+ (+ A C) (sqrt (fma (- A C) (- A C) (* B_m B_m))))) t_1))
           (if (<= t_4 INFINITY) t_2 (/ (sqrt (* 2.0 F)) (- (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(C, (A * -4.0), (B_m * B_m));
    	double t_1 = fma(B_m, B_m, (-4.0 * (A * C)));
    	double t_2 = sqrt(((F * t_0) * (2.0 * fma(2.0, C, (((B_m * B_m) * -0.5) / A))))) / -t_0;
    	double t_3 = (4.0 * A) * C;
    	double t_4 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_3) * F)) * ((A + C) + sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / (t_3 - pow(B_m, 2.0));
    	double tmp;
    	if (t_4 <= -((double) INFINITY)) {
    		tmp = t_2;
    	} else if (t_4 <= -2e-194) {
    		tmp = (sqrt((t_1 * (2.0 * F))) / -1.0) * (sqrt(((A + C) + sqrt(fma((A - C), (A - C), (B_m * B_m))))) / t_1);
    	} else if (t_4 <= ((double) INFINITY)) {
    		tmp = t_2;
    	} else {
    		tmp = sqrt((2.0 * F)) / -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(C, Float64(A * -4.0), Float64(B_m * B_m))
    	t_1 = fma(B_m, B_m, Float64(-4.0 * Float64(A * C)))
    	t_2 = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * fma(2.0, C, Float64(Float64(Float64(B_m * B_m) * -0.5) / A))))) / Float64(-t_0))
    	t_3 = Float64(Float64(4.0 * A) * C)
    	t_4 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_3) * F)) * Float64(Float64(A + C) + sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / Float64(t_3 - (B_m ^ 2.0)))
    	tmp = 0.0
    	if (t_4 <= Float64(-Inf))
    		tmp = t_2;
    	elseif (t_4 <= -2e-194)
    		tmp = Float64(Float64(sqrt(Float64(t_1 * Float64(2.0 * F))) / -1.0) * Float64(sqrt(Float64(Float64(A + C) + sqrt(fma(Float64(A - C), Float64(A - C), Float64(B_m * B_m))))) / t_1));
    	elseif (t_4 <= Inf)
    		tmp = t_2;
    	else
    		tmp = Float64(sqrt(Float64(2.0 * F)) / 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[(C * N[(A * -4.0), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$2 = N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * N[(2.0 * C + N[(N[(N[(B$95$m * B$95$m), $MachinePrecision] * -0.5), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision]}, Block[{t$95$3 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$4 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$3), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$3 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$4, (-Infinity)], t$95$2, If[LessEqual[t$95$4, -2e-194], N[(N[(N[Sqrt[N[(t$95$1 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / -1.0), $MachinePrecision] * N[(N[Sqrt[N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[(A - C), $MachinePrecision] * N[(A - C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / t$95$1), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$4, Infinity], t$95$2, N[(N[Sqrt[N[(2.0 * F), $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(C, A \cdot -4, B\_m \cdot B\_m\right)\\
    t_1 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\
    t_2 := \frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\
    t_3 := \left(4 \cdot A\right) \cdot C\\
    t_4 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_3\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_3 - {B\_m}^{2}}\\
    \mathbf{if}\;t\_4 \leq -\infty:\\
    \;\;\;\;t\_2\\
    
    \mathbf{elif}\;t\_4 \leq -2 \cdot 10^{-194}:\\
    \;\;\;\;\frac{\sqrt{t\_1 \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}}}{t\_1}\\
    
    \mathbf{elif}\;t\_4 \leq \infty:\\
    \;\;\;\;t\_2\\
    
    \mathbf{else}:\\
    \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 3 regimes
    2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -inf.0 or -2.00000000000000004e-194 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < +inf.0

      1. Initial program 9.3%

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

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

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

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

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

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

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

          \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        7. lower-*.f6427.2

          \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      5. Applied rewrites27.2%

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      6. Step-by-step derivation
        1. lift-/.f64N/A

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

          \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
      7. Applied rewrites27.2%

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

      if -inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -2.00000000000000004e-194

      1. Initial program 97.6%

        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. Add Preprocessing
      3. Applied rewrites99.3%

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

      if +inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)))

      1. Initial program 0.0%

        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      2. 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. lower-neg.f64N/A

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

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

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

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

          \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
        7. lower-/.f6416.4

          \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
      5. Applied rewrites16.4%

        \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
      6. Step-by-step derivation
        1. Applied rewrites25.2%

          \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
      7. Recombined 3 regimes into one program.
      8. Final simplification36.2%

        \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\frac{\sqrt{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(2 \cdot F\right)}}{-1} \cdot \frac{\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}}}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B}}\\ \end{array} \]
      9. Add Preprocessing

      Alternative 3: 54.2% accurate, 0.3× speedup?

      \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\ t_1 := \frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\ t_2 := \left(4 \cdot A\right) \cdot C\\ t_3 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_2\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_2 - {B\_m}^{2}}\\ \mathbf{if}\;t\_3 \leq -\infty:\\ \;\;\;\;t\_1\\ \mathbf{elif}\;t\_3 \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}} \cdot \left(\frac{-1}{t\_0} \cdot \sqrt{t\_0 \cdot \left(2 \cdot F\right)}\right)\\ \mathbf{elif}\;t\_3 \leq \infty:\\ \;\;\;\;t\_1\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
      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 C (* A -4.0) (* B_m B_m)))
              (t_1
               (/
                (sqrt (* (* F t_0) (* 2.0 (fma 2.0 C (/ (* (* B_m B_m) -0.5) A)))))
                (- t_0)))
              (t_2 (* (* 4.0 A) C))
              (t_3
               (/
                (sqrt
                 (*
                  (* 2.0 (* (- (pow B_m 2.0) t_2) F))
                  (+ (+ A C) (sqrt (+ (pow B_m 2.0) (pow (- A C) 2.0))))))
                (- t_2 (pow B_m 2.0)))))
         (if (<= t_3 (- INFINITY))
           t_1
           (if (<= t_3 -2e-194)
             (*
              (sqrt (+ (+ A C) (sqrt (fma (- A C) (- A C) (* B_m B_m)))))
              (* (/ -1.0 t_0) (sqrt (* t_0 (* 2.0 F)))))
             (if (<= t_3 INFINITY) t_1 (/ (sqrt (* 2.0 F)) (- (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(C, (A * -4.0), (B_m * B_m));
      	double t_1 = sqrt(((F * t_0) * (2.0 * fma(2.0, C, (((B_m * B_m) * -0.5) / A))))) / -t_0;
      	double t_2 = (4.0 * A) * C;
      	double t_3 = sqrt(((2.0 * ((pow(B_m, 2.0) - t_2) * F)) * ((A + C) + sqrt((pow(B_m, 2.0) + pow((A - C), 2.0)))))) / (t_2 - pow(B_m, 2.0));
      	double tmp;
      	if (t_3 <= -((double) INFINITY)) {
      		tmp = t_1;
      	} else if (t_3 <= -2e-194) {
      		tmp = sqrt(((A + C) + sqrt(fma((A - C), (A - C), (B_m * B_m))))) * ((-1.0 / t_0) * sqrt((t_0 * (2.0 * F))));
      	} else if (t_3 <= ((double) INFINITY)) {
      		tmp = t_1;
      	} else {
      		tmp = sqrt((2.0 * F)) / -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(C, Float64(A * -4.0), Float64(B_m * B_m))
      	t_1 = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * fma(2.0, C, Float64(Float64(Float64(B_m * B_m) * -0.5) / A))))) / Float64(-t_0))
      	t_2 = Float64(Float64(4.0 * A) * C)
      	t_3 = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_2) * F)) * Float64(Float64(A + C) + sqrt(Float64((B_m ^ 2.0) + (Float64(A - C) ^ 2.0)))))) / Float64(t_2 - (B_m ^ 2.0)))
      	tmp = 0.0
      	if (t_3 <= Float64(-Inf))
      		tmp = t_1;
      	elseif (t_3 <= -2e-194)
      		tmp = Float64(sqrt(Float64(Float64(A + C) + sqrt(fma(Float64(A - C), Float64(A - C), Float64(B_m * B_m))))) * Float64(Float64(-1.0 / t_0) * sqrt(Float64(t_0 * Float64(2.0 * F)))));
      	elseif (t_3 <= Inf)
      		tmp = t_1;
      	else
      		tmp = Float64(sqrt(Float64(2.0 * F)) / 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[(C * N[(A * -4.0), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * N[(2.0 * C + N[(N[(N[(B$95$m * B$95$m), $MachinePrecision] * -0.5), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision]}, Block[{t$95$2 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, Block[{t$95$3 = N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$2), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[Power[B$95$m, 2.0], $MachinePrecision] + N[Power[N[(A - C), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$2 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[t$95$3, (-Infinity)], t$95$1, If[LessEqual[t$95$3, -2e-194], N[(N[Sqrt[N[(N[(A + C), $MachinePrecision] + N[Sqrt[N[(N[(A - C), $MachinePrecision] * N[(A - C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(-1.0 / t$95$0), $MachinePrecision] * N[Sqrt[N[(t$95$0 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$3, Infinity], t$95$1, N[(N[Sqrt[N[(2.0 * F), $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(C, A \cdot -4, B\_m \cdot B\_m\right)\\
      t_1 := \frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\
      t_2 := \left(4 \cdot A\right) \cdot C\\
      t_3 := \frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_2\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B\_m}^{2} + {\left(A - C\right)}^{2}}\right)}}{t\_2 - {B\_m}^{2}}\\
      \mathbf{if}\;t\_3 \leq -\infty:\\
      \;\;\;\;t\_1\\
      
      \mathbf{elif}\;t\_3 \leq -2 \cdot 10^{-194}:\\
      \;\;\;\;\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B\_m \cdot B\_m\right)}} \cdot \left(\frac{-1}{t\_0} \cdot \sqrt{t\_0 \cdot \left(2 \cdot F\right)}\right)\\
      
      \mathbf{elif}\;t\_3 \leq \infty:\\
      \;\;\;\;t\_1\\
      
      \mathbf{else}:\\
      \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B\_m}}\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 3 regimes
      2. if (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -inf.0 or -2.00000000000000004e-194 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < +inf.0

        1. Initial program 9.3%

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

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

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

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

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

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

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

            \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          7. lower-*.f6427.2

            \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        5. Applied rewrites27.2%

          \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        6. Step-by-step derivation
          1. lift-/.f64N/A

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

            \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
        7. Applied rewrites27.2%

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

        if -inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C))) < -2.00000000000000004e-194

        1. Initial program 97.6%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. Add Preprocessing
        3. Applied rewrites97.4%

          \[\leadsto \color{blue}{\sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(\sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)} + \left(A + C\right)\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)}} \]
        4. Applied rewrites99.1%

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

        if +inf.0 < (/.f64 (neg.f64 (sqrt.f64 (*.f64 (*.f64 #s(literal 2 binary64) (*.f64 (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)) F)) (+.f64 (+.f64 A C) (sqrt.f64 (+.f64 (pow.f64 (-.f64 A C) #s(literal 2 binary64)) (pow.f64 B #s(literal 2 binary64)))))))) (-.f64 (pow.f64 B #s(literal 2 binary64)) (*.f64 (*.f64 #s(literal 4 binary64) A) C)))

        1. Initial program 0.0%

          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        2. 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. lower-neg.f64N/A

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

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

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

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

            \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
          7. lower-/.f6416.4

            \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
        5. Applied rewrites16.4%

          \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
        6. Step-by-step derivation
          1. Applied rewrites25.2%

            \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
        7. Recombined 3 regimes into one program.
        8. Final simplification36.2%

          \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -\infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq -2 \cdot 10^{-194}:\\ \;\;\;\;\sqrt{\left(A + C\right) + \sqrt{\mathsf{fma}\left(A - C, A - C, B \cdot B\right)}} \cdot \left(\frac{-1}{\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)} \cdot \sqrt{\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right) \cdot \left(2 \cdot F\right)}\right)\\ \mathbf{elif}\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}} \leq \infty:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B}}\\ \end{array} \]
        9. Add Preprocessing

        Alternative 4: 46.5% accurate, 1.7× speedup?

        \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \sqrt{2 \cdot F}\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-186}:\\ \;\;\;\;\frac{-1}{\mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)} \cdot \sqrt{-16 \cdot \left(F \cdot \left(A \cdot \left(C \cdot C\right)\right)\right)}\\ \mathbf{elif}\;{B\_m}^{2} \leq 10^{+252}:\\ \;\;\;\;\frac{t\_0 \cdot \sqrt{C + \sqrt{\mathsf{fma}\left(B\_m, B\_m, C \cdot C\right)}}}{-B\_m}\\ \mathbf{else}:\\ \;\;\;\;\frac{t\_0}{-\sqrt{B\_m}}\\ \end{array} \end{array} \]
        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 (sqrt (* 2.0 F))))
           (if (<= (pow B_m 2.0) 2e-186)
             (*
              (/ -1.0 (fma B_m B_m (* -4.0 (* A C))))
              (sqrt (* -16.0 (* F (* A (* C C))))))
             (if (<= (pow B_m 2.0) 1e+252)
               (/ (* t_0 (sqrt (+ C (sqrt (fma B_m B_m (* C C)))))) (- B_m))
               (/ t_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 = sqrt((2.0 * F));
        	double tmp;
        	if (pow(B_m, 2.0) <= 2e-186) {
        		tmp = (-1.0 / fma(B_m, B_m, (-4.0 * (A * C)))) * sqrt((-16.0 * (F * (A * (C * C)))));
        	} else if (pow(B_m, 2.0) <= 1e+252) {
        		tmp = (t_0 * sqrt((C + sqrt(fma(B_m, B_m, (C * C)))))) / -B_m;
        	} else {
        		tmp = t_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 = sqrt(Float64(2.0 * F))
        	tmp = 0.0
        	if ((B_m ^ 2.0) <= 2e-186)
        		tmp = Float64(Float64(-1.0 / fma(B_m, B_m, Float64(-4.0 * Float64(A * C)))) * sqrt(Float64(-16.0 * Float64(F * Float64(A * Float64(C * C))))));
        	elseif ((B_m ^ 2.0) <= 1e+252)
        		tmp = Float64(Float64(t_0 * sqrt(Float64(C + sqrt(fma(B_m, B_m, Float64(C * C)))))) / Float64(-B_m));
        	else
        		tmp = Float64(t_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[Sqrt[N[(2.0 * F), $MachinePrecision]], $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-186], N[(N[(-1.0 / N[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(-16.0 * N[(F * N[(A * N[(C * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e+252], N[(N[(t$95$0 * N[Sqrt[N[(C + N[Sqrt[N[(B$95$m * B$95$m + N[(C * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / (-B$95$m)), $MachinePrecision], N[(t$95$0 / (-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 := \sqrt{2 \cdot F}\\
        \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-186}:\\
        \;\;\;\;\frac{-1}{\mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)} \cdot \sqrt{-16 \cdot \left(F \cdot \left(A \cdot \left(C \cdot C\right)\right)\right)}\\
        
        \mathbf{elif}\;{B\_m}^{2} \leq 10^{+252}:\\
        \;\;\;\;\frac{t\_0 \cdot \sqrt{C + \sqrt{\mathsf{fma}\left(B\_m, B\_m, C \cdot C\right)}}}{-B\_m}\\
        
        \mathbf{else}:\\
        \;\;\;\;\frac{t\_0}{-\sqrt{B\_m}}\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 3 regimes
        2. if (pow.f64 B #s(literal 2 binary64)) < 1.9999999999999998e-186

          1. Initial program 18.6%

            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          2. Add Preprocessing
          3. Applied rewrites18.4%

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

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

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

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

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

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

              \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
            6. lower-*.f649.0

              \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
          6. Applied rewrites9.0%

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

          if 1.9999999999999998e-186 < (pow.f64 B #s(literal 2 binary64)) < 1.0000000000000001e252

          1. Initial program 27.6%

            \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          2. Add Preprocessing
          3. Taylor expanded in A around 0

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

              \[\leadsto \color{blue}{\mathsf{neg}\left(\frac{\sqrt{2}}{B} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}\right)} \]
            2. associate-*l/N/A

              \[\leadsto \mathsf{neg}\left(\color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}{B}}\right) \]
            3. distribute-neg-frac2N/A

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

              \[\leadsto \frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}{\color{blue}{-1 \cdot B}} \]
            5. lower-/.f64N/A

              \[\leadsto \color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C + \sqrt{{B}^{2} + {C}^{2}}\right)}}{-1 \cdot B}} \]
          5. Applied rewrites19.7%

            \[\leadsto \color{blue}{\frac{\sqrt{2} \cdot \sqrt{F \cdot \left(C + \sqrt{\mathsf{fma}\left(C, C, B \cdot B\right)}\right)}}{-B}} \]
          6. Step-by-step derivation
            1. Applied rewrites20.1%

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

            if 1.0000000000000001e252 < (pow.f64 B #s(literal 2 binary64))

            1. Initial program 3.1%

              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            2. 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. lower-neg.f64N/A

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

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

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

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

                \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
              7. lower-/.f6426.3

                \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
            5. Applied rewrites26.3%

              \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
            6. Step-by-step derivation
              1. Applied rewrites39.9%

                \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
            7. Recombined 3 regimes into one program.
            8. Final simplification21.4%

              \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-186}:\\ \;\;\;\;\frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \cdot \sqrt{-16 \cdot \left(F \cdot \left(A \cdot \left(C \cdot C\right)\right)\right)}\\ \mathbf{elif}\;{B}^{2} \leq 10^{+252}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F} \cdot \sqrt{C + \sqrt{\mathsf{fma}\left(B, B, C \cdot C\right)}}}{-B}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2 \cdot F}}{-\sqrt{B}}\\ \end{array} \]
            9. Add Preprocessing

            Alternative 5: 52.2% accurate, 2.4× speedup?

            \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+46}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}\\ \end{array} \end{array} \]
            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 C (* A -4.0) (* B_m B_m))))
               (if (<= (pow B_m 2.0) 2e+46)
                 (/
                  (sqrt (* (* F t_0) (* 2.0 (fma 2.0 C (/ (* (* B_m B_m) -0.5) A)))))
                  (- t_0))
                 (- (* (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) {
            	double t_0 = fma(C, (A * -4.0), (B_m * B_m));
            	double tmp;
            	if (pow(B_m, 2.0) <= 2e+46) {
            		tmp = sqrt(((F * t_0) * (2.0 * fma(2.0, C, (((B_m * B_m) * -0.5) / A))))) / -t_0;
            	} else {
            		tmp = -(sqrt(F) * sqrt((2.0 / 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(C, Float64(A * -4.0), Float64(B_m * B_m))
            	tmp = 0.0
            	if ((B_m ^ 2.0) <= 2e+46)
            		tmp = Float64(sqrt(Float64(Float64(F * t_0) * Float64(2.0 * fma(2.0, C, Float64(Float64(Float64(B_m * B_m) * -0.5) / A))))) / Float64(-t_0));
            	else
            		tmp = Float64(-Float64(sqrt(F) * sqrt(Float64(2.0 / 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[(C * N[(A * -4.0), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e+46], N[(N[Sqrt[N[(N[(F * t$95$0), $MachinePrecision] * N[(2.0 * N[(2.0 * C + N[(N[(N[(B$95$m * B$95$m), $MachinePrecision] * -0.5), $MachinePrecision] / A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], (-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])\\
            \\
            \begin{array}{l}
            t_0 := \mathsf{fma}\left(C, A \cdot -4, B\_m \cdot B\_m\right)\\
            \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{+46}:\\
            \;\;\;\;\frac{\sqrt{\left(F \cdot t\_0\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B\_m \cdot B\_m\right) \cdot -0.5}{A}\right)\right)}}{-t\_0}\\
            
            \mathbf{else}:\\
            \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}\\
            
            
            \end{array}
            \end{array}
            
            Derivation
            1. Split input into 2 regimes
            2. if (pow.f64 B #s(literal 2 binary64)) < 2e46

              1. Initial program 22.0%

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

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

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

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

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

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

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

                  \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                7. lower-*.f6423.2

                  \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{-0.5 \cdot \color{blue}{\left(B \cdot B\right)}}{A}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              5. Applied rewrites23.2%

                \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\mathsf{fma}\left(2, C, \frac{-0.5 \cdot \left(B \cdot B\right)}{A}\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              6. Step-by-step derivation
                1. lift-/.f64N/A

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

                  \[\leadsto \color{blue}{\frac{\mathsf{neg}\left(\left(\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \mathsf{fma}\left(2, C, \frac{\frac{-1}{2} \cdot \left(B \cdot B\right)}{A}\right)}\right)\right)\right)}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
              7. Applied rewrites23.2%

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

              if 2e46 < (pow.f64 B #s(literal 2 binary64))

              1. Initial program 11.4%

                \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              2. 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. lower-neg.f64N/A

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

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

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

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

                  \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                7. lower-/.f6425.2

                  \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
              5. Applied rewrites25.2%

                \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
              6. Step-by-step derivation
                1. Applied rewrites33.7%

                  \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
                2. Step-by-step derivation
                  1. Applied rewrites33.7%

                    \[\leadsto -\sqrt{F} \cdot \sqrt{\frac{2}{B}} \]
                3. Recombined 2 regimes into one program.
                4. Final simplification27.8%

                  \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{+46}:\\ \;\;\;\;\frac{\sqrt{\left(F \cdot \mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)\right) \cdot \left(2 \cdot \mathsf{fma}\left(2, C, \frac{\left(B \cdot B\right) \cdot -0.5}{A}\right)\right)}}{-\mathsf{fma}\left(C, A \cdot -4, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B}}\\ \end{array} \]
                5. Add Preprocessing

                Alternative 6: 52.3% accurate, 2.6× speedup?

                \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-11}:\\ \;\;\;\;\sqrt{\left(t\_0 \cdot \left(2 \cdot F\right)\right) \cdot \left(2 \cdot C\right)} \cdot \frac{-1}{t\_0}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}\\ \end{array} \end{array} \]
                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 B_m B_m (* -4.0 (* A C)))))
                   (if (<= (pow B_m 2.0) 1e-11)
                     (* (sqrt (* (* t_0 (* 2.0 F)) (* 2.0 C))) (/ -1.0 t_0))
                     (- (* (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) {
                	double t_0 = fma(B_m, B_m, (-4.0 * (A * C)));
                	double tmp;
                	if (pow(B_m, 2.0) <= 1e-11) {
                		tmp = sqrt(((t_0 * (2.0 * F)) * (2.0 * C))) * (-1.0 / t_0);
                	} else {
                		tmp = -(sqrt(F) * sqrt((2.0 / 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(B_m, B_m, Float64(-4.0 * Float64(A * C)))
                	tmp = 0.0
                	if ((B_m ^ 2.0) <= 1e-11)
                		tmp = Float64(sqrt(Float64(Float64(t_0 * Float64(2.0 * F)) * Float64(2.0 * C))) * Float64(-1.0 / t_0));
                	else
                		tmp = Float64(-Float64(sqrt(F) * sqrt(Float64(2.0 / 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[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-11], N[(N[Sqrt[N[(N[(t$95$0 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision] * N[(2.0 * C), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / t$95$0), $MachinePrecision]), $MachinePrecision], (-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])\\
                \\
                \begin{array}{l}
                t_0 := \mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)\\
                \mathbf{if}\;{B\_m}^{2} \leq 10^{-11}:\\
                \;\;\;\;\sqrt{\left(t\_0 \cdot \left(2 \cdot F\right)\right) \cdot \left(2 \cdot C\right)} \cdot \frac{-1}{t\_0}\\
                
                \mathbf{else}:\\
                \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}\\
                
                
                \end{array}
                \end{array}
                
                Derivation
                1. Split input into 2 regimes
                2. if (pow.f64 B #s(literal 2 binary64)) < 9.99999999999999939e-12

                  1. Initial program 19.5%

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

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

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

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

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

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

                      \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \left(C \cdot \color{blue}{2}\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
                    5. lower-*.f6418.9

                      \[\leadsto \sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(F \cdot 2\right)\right) \cdot \color{blue}{\left(C \cdot 2\right)}} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
                  6. Applied rewrites18.9%

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

                  if 9.99999999999999939e-12 < (pow.f64 B #s(literal 2 binary64))

                  1. Initial program 15.1%

                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                  2. 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. lower-neg.f64N/A

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

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

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

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

                      \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                    7. lower-/.f6425.2

                      \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                  5. Applied rewrites25.2%

                    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                  6. Step-by-step derivation
                    1. Applied rewrites32.8%

                      \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
                    2. Step-by-step derivation
                      1. Applied rewrites32.8%

                        \[\leadsto -\sqrt{F} \cdot \sqrt{\frac{2}{B}} \]
                    3. Recombined 2 regimes into one program.
                    4. Final simplification25.7%

                      \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-11}:\\ \;\;\;\;\sqrt{\left(\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right) \cdot \left(2 \cdot F\right)\right) \cdot \left(2 \cdot C\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B}}\\ \end{array} \]
                    5. Add Preprocessing

                    Alternative 7: 44.7% accurate, 2.9× speedup?

                    \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-46}:\\ \;\;\;\;\frac{-1}{\mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)} \cdot \sqrt{-16 \cdot \left(F \cdot \left(A \cdot \left(C \cdot C\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}\\ \end{array} \end{array} \]
                    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 (<= (pow B_m 2.0) 5e-46)
                       (*
                        (/ -1.0 (fma B_m B_m (* -4.0 (* A C))))
                        (sqrt (* -16.0 (* F (* A (* C C))))))
                       (- (* (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) {
                    	double tmp;
                    	if (pow(B_m, 2.0) <= 5e-46) {
                    		tmp = (-1.0 / fma(B_m, B_m, (-4.0 * (A * C)))) * sqrt((-16.0 * (F * (A * (C * C)))));
                    	} else {
                    		tmp = -(sqrt(F) * sqrt((2.0 / 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 ^ 2.0) <= 5e-46)
                    		tmp = Float64(Float64(-1.0 / fma(B_m, B_m, Float64(-4.0 * Float64(A * C)))) * sqrt(Float64(-16.0 * Float64(F * Float64(A * Float64(C * C))))));
                    	else
                    		tmp = Float64(-Float64(sqrt(F) * sqrt(Float64(2.0 / 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[N[Power[B$95$m, 2.0], $MachinePrecision], 5e-46], N[(N[(-1.0 / N[(B$95$m * B$95$m + N[(-4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(-16.0 * N[(F * N[(A * N[(C * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], (-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])\\
                    \\
                    \begin{array}{l}
                    \mathbf{if}\;{B\_m}^{2} \leq 5 \cdot 10^{-46}:\\
                    \;\;\;\;\frac{-1}{\mathsf{fma}\left(B\_m, B\_m, -4 \cdot \left(A \cdot C\right)\right)} \cdot \sqrt{-16 \cdot \left(F \cdot \left(A \cdot \left(C \cdot C\right)\right)\right)}\\
                    
                    \mathbf{else}:\\
                    \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}\\
                    
                    
                    \end{array}
                    \end{array}
                    
                    Derivation
                    1. Split input into 2 regimes
                    2. if (pow.f64 B #s(literal 2 binary64)) < 4.99999999999999992e-46

                      1. Initial program 17.9%

                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                      2. Add Preprocessing
                      3. Applied rewrites17.8%

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

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

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

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

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

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

                          \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
                        6. lower-*.f6410.3

                          \[\leadsto \sqrt{-16 \cdot \left(\left(A \cdot \color{blue}{\left(C \cdot C\right)}\right) \cdot F\right)} \cdot \frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \]
                      6. Applied rewrites10.3%

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

                      if 4.99999999999999992e-46 < (pow.f64 B #s(literal 2 binary64))

                      1. Initial program 16.9%

                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                      2. 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. lower-neg.f64N/A

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

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

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

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

                          \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                        7. lower-/.f6424.7

                          \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                      5. Applied rewrites24.7%

                        \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                      6. Step-by-step derivation
                        1. Applied rewrites31.7%

                          \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
                        2. Step-by-step derivation
                          1. Applied rewrites31.7%

                            \[\leadsto -\sqrt{F} \cdot \sqrt{\frac{2}{B}} \]
                        3. Recombined 2 regimes into one program.
                        4. Final simplification21.7%

                          \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 5 \cdot 10^{-46}:\\ \;\;\;\;\frac{-1}{\mathsf{fma}\left(B, B, -4 \cdot \left(A \cdot C\right)\right)} \cdot \sqrt{-16 \cdot \left(F \cdot \left(A \cdot \left(C \cdot C\right)\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;-\sqrt{F} \cdot \sqrt{\frac{2}{B}}\\ \end{array} \]
                        5. Add Preprocessing

                        Alternative 8: 35.4% accurate, 12.6× speedup?

                        \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}} \end{array} \]
                        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])\\
                        \\
                        -\sqrt{F} \cdot \sqrt{\frac{2}{B\_m}}
                        \end{array}
                        
                        Derivation
                        1. Initial program 17.4%

                          \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                        2. 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. lower-neg.f64N/A

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

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

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

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

                            \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                          7. lower-/.f6414.9

                            \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                        5. Applied rewrites14.9%

                          \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                        6. Step-by-step derivation
                          1. Applied rewrites19.0%

                            \[\leadsto -\frac{\sqrt{2 \cdot F}}{\sqrt{B}} \]
                          2. Step-by-step derivation
                            1. Applied rewrites19.0%

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

                            Alternative 9: 27.2% accurate, 15.8× speedup?

                            \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{\left|2 \cdot \frac{F}{B\_m}\right|} \end{array} \]
                            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 (fabs (* 2.0 (/ F 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(fabs((2.0 * (F / 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(abs((2.0d0 * (f / 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(Math.abs((2.0 * (F / 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(math.fabs((2.0 * (F / 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(abs(Float64(2.0 * Float64(F / 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(abs((2.0 * (F / 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[Sqrt[N[Abs[N[(2.0 * N[(F / B$95$m), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision])
                            
                            \begin{array}{l}
                            B_m = \left|B\right|
                            \\
                            [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                            \\
                            -\sqrt{\left|2 \cdot \frac{F}{B\_m}\right|}
                            \end{array}
                            
                            Derivation
                            1. Initial program 17.4%

                              \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                            2. 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. lower-neg.f64N/A

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

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

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

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

                                \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                              7. lower-/.f6414.9

                                \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                            5. Applied rewrites14.9%

                              \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                            6. Step-by-step derivation
                              1. Applied rewrites15.0%

                                \[\leadsto \color{blue}{-\sqrt{\frac{2 \cdot F}{B}}} \]
                              2. Step-by-step derivation
                                1. Applied rewrites27.7%

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

                                Alternative 10: 27.1% accurate, 16.9× speedup?

                                \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{\frac{2 \cdot F}{B\_m}} \end{array} \]
                                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 (/ (* 2.0 F) 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(((2.0 * F) / 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(((2.0d0 * f) / 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(((2.0 * F) / 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(((2.0 * F) / 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(Float64(2.0 * F) / 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(((2.0 * F) / 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[Sqrt[N[(N[(2.0 * F), $MachinePrecision] / 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])\\
                                \\
                                -\sqrt{\frac{2 \cdot F}{B\_m}}
                                \end{array}
                                
                                Derivation
                                1. Initial program 17.4%

                                  \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                2. 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. lower-neg.f64N/A

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

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

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

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

                                    \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                                  7. lower-/.f6414.9

                                    \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                5. Applied rewrites14.9%

                                  \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                                6. Step-by-step derivation
                                  1. Applied rewrites15.0%

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

                                  Alternative 11: 27.1% accurate, 16.9× speedup?

                                  \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{2 \cdot \frac{F}{B\_m}} \end{array} \]
                                  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 (* 2.0 (/ F 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((2.0 * (F / 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((2.0d0 * (f / 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((2.0 * (F / 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((2.0 * (F / 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(2.0 * Float64(F / 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((2.0 * (F / 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[Sqrt[N[(2.0 * N[(F / 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])\\
                                  \\
                                  -\sqrt{2 \cdot \frac{F}{B\_m}}
                                  \end{array}
                                  
                                  Derivation
                                  1. Initial program 17.4%

                                    \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                  2. 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. lower-neg.f64N/A

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

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

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

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

                                      \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                                    7. lower-/.f6414.9

                                      \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                  5. Applied rewrites14.9%

                                    \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                                  6. Step-by-step derivation
                                    1. Applied rewrites15.0%

                                      \[\leadsto \color{blue}{-\sqrt{\frac{2 \cdot F}{B}}} \]
                                    2. Step-by-step derivation
                                      1. Applied rewrites15.0%

                                        \[\leadsto -\sqrt{\frac{F}{B} \cdot 2} \]
                                      2. Final simplification15.0%

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

                                      Alternative 12: 27.1% accurate, 16.9× speedup?

                                      \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ -\sqrt{F \cdot \frac{2}{B\_m}} \end{array} \]
                                      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 (/ 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 * (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 * (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 * (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 * (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(-sqrt(Float64(F * 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 * (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[Sqrt[N[(F * 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])\\
                                      \\
                                      -\sqrt{F \cdot \frac{2}{B\_m}}
                                      \end{array}
                                      
                                      Derivation
                                      1. Initial program 17.4%

                                        \[\frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(\left(A + C\right) + \sqrt{{\left(A - C\right)}^{2} + {B}^{2}}\right)}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                                      2. 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. lower-neg.f64N/A

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

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

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

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

                                          \[\leadsto \mathsf{neg}\left(\sqrt{2} \cdot \color{blue}{\sqrt{\frac{F}{B}}}\right) \]
                                        7. lower-/.f6414.9

                                          \[\leadsto -\sqrt{2} \cdot \sqrt{\color{blue}{\frac{F}{B}}} \]
                                      5. Applied rewrites14.9%

                                        \[\leadsto \color{blue}{-\sqrt{2} \cdot \sqrt{\frac{F}{B}}} \]
                                      6. Step-by-step derivation
                                        1. Applied rewrites15.0%

                                          \[\leadsto \color{blue}{-\sqrt{\frac{2 \cdot F}{B}}} \]
                                        2. Step-by-step derivation
                                          1. Applied rewrites15.0%

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

                                          Reproduce

                                          ?
                                          herbie shell --seed 2024219 
                                          (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))))