ABCF->ab-angle b

Percentage Accurate: 18.4% → 43.2%
Time: 17.6s
Alternatives: 10
Speedup: 12.0×

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 10 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: 18.4% 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: 43.2% accurate, 1.3× speedup?

\[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \left(4 \cdot A\right) \cdot C\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{t\_0 - {B\_m}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
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 (* (* 4.0 A) C)))
   (if (<= (pow B_m 2.0) 2e-31)
     (/
      (sqrt (* (* 2.0 (* (- (pow B_m 2.0) t_0) F)) (+ A A)))
      (- t_0 (pow B_m 2.0)))
     (* (sqrt (/ F (- B_m))) (- (sqrt 2.0))))))
B_m = fabs(B);
assert(A < B_m && B_m < C && C < F);
double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (pow(B_m, 2.0) <= 2e-31) {
		tmp = sqrt(((2.0 * ((pow(B_m, 2.0) - t_0) * F)) * (A + A))) / (t_0 - pow(B_m, 2.0));
	} else {
		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
	}
	return tmp;
}
B_m = abs(b)
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
real(8) function code(a, b_m, c, f)
    real(8), intent (in) :: a
    real(8), intent (in) :: b_m
    real(8), intent (in) :: c
    real(8), intent (in) :: f
    real(8) :: t_0
    real(8) :: tmp
    t_0 = (4.0d0 * a) * c
    if ((b_m ** 2.0d0) <= 2d-31) then
        tmp = sqrt(((2.0d0 * (((b_m ** 2.0d0) - t_0) * f)) * (a + a))) / (t_0 - (b_m ** 2.0d0))
    else
        tmp = sqrt((f / -b_m)) * -sqrt(2.0d0)
    end if
    code = tmp
end function
B_m = Math.abs(B);
assert A < B_m && B_m < C && C < F;
public static double code(double A, double B_m, double C, double F) {
	double t_0 = (4.0 * A) * C;
	double tmp;
	if (Math.pow(B_m, 2.0) <= 2e-31) {
		tmp = Math.sqrt(((2.0 * ((Math.pow(B_m, 2.0) - t_0) * F)) * (A + A))) / (t_0 - Math.pow(B_m, 2.0));
	} else {
		tmp = Math.sqrt((F / -B_m)) * -Math.sqrt(2.0);
	}
	return tmp;
}
B_m = math.fabs(B)
[A, B_m, C, F] = sort([A, B_m, C, F])
def code(A, B_m, C, F):
	t_0 = (4.0 * A) * C
	tmp = 0
	if math.pow(B_m, 2.0) <= 2e-31:
		tmp = math.sqrt(((2.0 * ((math.pow(B_m, 2.0) - t_0) * F)) * (A + A))) / (t_0 - math.pow(B_m, 2.0))
	else:
		tmp = math.sqrt((F / -B_m)) * -math.sqrt(2.0)
	return tmp
B_m = abs(B)
A, B_m, C, F = sort([A, B_m, C, F])
function code(A, B_m, C, F)
	t_0 = Float64(Float64(4.0 * A) * C)
	tmp = 0.0
	if ((B_m ^ 2.0) <= 2e-31)
		tmp = Float64(sqrt(Float64(Float64(2.0 * Float64(Float64((B_m ^ 2.0) - t_0) * F)) * Float64(A + A))) / Float64(t_0 - (B_m ^ 2.0)));
	else
		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)));
	end
	return tmp
end
B_m = abs(B);
A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
function tmp_2 = code(A, B_m, C, F)
	t_0 = (4.0 * A) * C;
	tmp = 0.0;
	if ((B_m ^ 2.0) <= 2e-31)
		tmp = sqrt(((2.0 * (((B_m ^ 2.0) - t_0) * F)) * (A + A))) / (t_0 - (B_m ^ 2.0));
	else
		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
	end
	tmp_2 = tmp;
end
B_m = N[Abs[B], $MachinePrecision]
NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(N[(4.0 * A), $MachinePrecision] * C), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-31], N[(N[Sqrt[N[(N[(2.0 * N[(N[(N[Power[B$95$m, 2.0], $MachinePrecision] - t$95$0), $MachinePrecision] * F), $MachinePrecision]), $MachinePrecision] * N[(A + A), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(t$95$0 - N[Power[B$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $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 := \left(4 \cdot A\right) \cdot C\\
\mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\
\;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B\_m}^{2} - t\_0\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{t\_0 - {B\_m}^{2}}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\


\end{array}
\end{array}
Derivation
  1. Split input into 2 regimes
  2. if (pow.f64 B #s(literal 2 binary64)) < 2e-31

    1. Initial program 14.7%

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

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

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

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

        \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + \color{blue}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      4. lower-+.f6426.1

        \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(A + A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
    5. Applied rewrites26.1%

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

    if 2e-31 < (pow.f64 B #s(literal 2 binary64))

    1. Initial program 16.4%

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

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

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

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

        \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
    5. Applied rewrites20.8%

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

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

        \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
    8. Recombined 2 regimes into one program.
    9. Final simplification27.1%

      \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{\left(4 \cdot A\right) \cdot C - {B}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
    10. Add Preprocessing

    Alternative 2: 43.2% accurate, 2.6× speedup?

    \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(A + A\right) \cdot \left(t\_0 \cdot \left(2 \cdot F\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
    B_m = (fabs.f64 B)
    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
    (FPCore (A B_m C F)
     :precision binary64
     (let* ((t_0 (fma -4.0 (* A C) (* B_m B_m))))
       (if (<= (pow B_m 2.0) 2e-31)
         (/ -1.0 (/ t_0 (sqrt (* (+ A A) (* t_0 (* 2.0 F))))))
         (* (sqrt (/ F (- B_m))) (- (sqrt 2.0))))))
    B_m = fabs(B);
    assert(A < B_m && B_m < C && C < F);
    double code(double A, double B_m, double C, double F) {
    	double t_0 = fma(-4.0, (A * C), (B_m * B_m));
    	double tmp;
    	if (pow(B_m, 2.0) <= 2e-31) {
    		tmp = -1.0 / (t_0 / sqrt(((A + A) * (t_0 * (2.0 * F)))));
    	} else {
    		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
    	}
    	return tmp;
    }
    
    B_m = abs(B)
    A, B_m, C, F = sort([A, B_m, C, F])
    function code(A, B_m, C, F)
    	t_0 = fma(-4.0, Float64(A * C), Float64(B_m * B_m))
    	tmp = 0.0
    	if ((B_m ^ 2.0) <= 2e-31)
    		tmp = Float64(-1.0 / Float64(t_0 / sqrt(Float64(Float64(A + A) * Float64(t_0 * Float64(2.0 * F))))));
    	else
    		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)));
    	end
    	return tmp
    end
    
    B_m = N[Abs[B], $MachinePrecision]
    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
    code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(-4.0 * N[(A * C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-31], N[(-1.0 / N[(t$95$0 / N[Sqrt[N[(N[(A + A), $MachinePrecision] * N[(t$95$0 * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision]]]
    
    \begin{array}{l}
    B_m = \left|B\right|
    \\
    [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
    \\
    \begin{array}{l}
    t_0 := \mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)\\
    \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\
    \;\;\;\;\frac{-1}{\frac{t\_0}{\sqrt{\left(A + A\right) \cdot \left(t\_0 \cdot \left(2 \cdot F\right)\right)}}}\\
    
    \mathbf{else}:\\
    \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\
    
    
    \end{array}
    \end{array}
    
    Derivation
    1. Split input into 2 regimes
    2. if (pow.f64 B #s(literal 2 binary64)) < 2e-31

      1. Initial program 14.7%

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

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

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

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

          \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + \color{blue}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        4. lower-+.f6426.1

          \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(A + A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
      5. Applied rewrites26.1%

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

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

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

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

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

          \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
      7. Applied rewrites26.1%

        \[\leadsto \color{blue}{\frac{\sqrt{2 \cdot \left(\left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)}} \]
      8. Applied rewrites26.1%

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

      if 2e-31 < (pow.f64 B #s(literal 2 binary64))

      1. Initial program 16.4%

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

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

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

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

          \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
      5. Applied rewrites20.8%

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

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

          \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
      8. Recombined 2 regimes into one program.
      9. Final simplification27.1%

        \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{-1}{\frac{\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)}{\sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot \left(2 \cdot F\right)\right)}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
      10. Add Preprocessing

      Alternative 3: 43.2% accurate, 2.7× speedup?

      \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := \mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)\\ \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(F \cdot t\_0\right)\right)}}{-t\_0}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
      B_m = (fabs.f64 B)
      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
      (FPCore (A B_m C F)
       :precision binary64
       (let* ((t_0 (fma -4.0 (* A C) (* B_m B_m))))
         (if (<= (pow B_m 2.0) 2e-31)
           (/ (sqrt (* 2.0 (* (+ A A) (* F t_0)))) (- t_0))
           (* (sqrt (/ F (- B_m))) (- (sqrt 2.0))))))
      B_m = fabs(B);
      assert(A < B_m && B_m < C && C < F);
      double code(double A, double B_m, double C, double F) {
      	double t_0 = fma(-4.0, (A * C), (B_m * B_m));
      	double tmp;
      	if (pow(B_m, 2.0) <= 2e-31) {
      		tmp = sqrt((2.0 * ((A + A) * (F * t_0)))) / -t_0;
      	} else {
      		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
      	}
      	return tmp;
      }
      
      B_m = abs(B)
      A, B_m, C, F = sort([A, B_m, C, F])
      function code(A, B_m, C, F)
      	t_0 = fma(-4.0, Float64(A * C), Float64(B_m * B_m))
      	tmp = 0.0
      	if ((B_m ^ 2.0) <= 2e-31)
      		tmp = Float64(sqrt(Float64(2.0 * Float64(Float64(A + A) * Float64(F * t_0)))) / Float64(-t_0));
      	else
      		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)));
      	end
      	return tmp
      end
      
      B_m = N[Abs[B], $MachinePrecision]
      NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
      code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = N[(-4.0 * N[(A * C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 2e-31], N[(N[Sqrt[N[(2.0 * N[(N[(A + A), $MachinePrecision] * N[(F * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-t$95$0)), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision]]]
      
      \begin{array}{l}
      B_m = \left|B\right|
      \\
      [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
      \\
      \begin{array}{l}
      t_0 := \mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)\\
      \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\
      \;\;\;\;\frac{\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(F \cdot t\_0\right)\right)}}{-t\_0}\\
      
      \mathbf{else}:\\
      \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\
      
      
      \end{array}
      \end{array}
      
      Derivation
      1. Split input into 2 regimes
      2. if (pow.f64 B #s(literal 2 binary64)) < 2e-31

        1. Initial program 14.7%

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

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

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

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

            \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + \color{blue}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          4. lower-+.f6426.1

            \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(A + A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
        5. Applied rewrites26.1%

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

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

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

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

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

            \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
        7. Applied rewrites26.1%

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

        if 2e-31 < (pow.f64 B #s(literal 2 binary64))

        1. Initial program 16.4%

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

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

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

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

            \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
        5. Applied rewrites20.8%

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

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

            \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
        8. Recombined 2 regimes into one program.
        9. Final simplification27.1%

          \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(F \cdot \mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)\right)\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
        10. Add Preprocessing

        Alternative 4: 42.3% accurate, 2.8× speedup?

        \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(F \cdot \mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
        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) 2e-31)
           (/
            (sqrt (* 2.0 (* (+ A A) (* F (fma -4.0 (* A C) (* B_m B_m))))))
            (* 4.0 (* A C)))
           (* (sqrt (/ F (- B_m))) (- (sqrt 2.0)))))
        B_m = fabs(B);
        assert(A < B_m && B_m < C && C < F);
        double code(double A, double B_m, double C, double F) {
        	double tmp;
        	if (pow(B_m, 2.0) <= 2e-31) {
        		tmp = sqrt((2.0 * ((A + A) * (F * fma(-4.0, (A * C), (B_m * B_m)))))) / (4.0 * (A * C));
        	} else {
        		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
        	}
        	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) <= 2e-31)
        		tmp = Float64(sqrt(Float64(2.0 * Float64(Float64(A + A) * Float64(F * fma(-4.0, Float64(A * C), Float64(B_m * B_m)))))) / Float64(4.0 * Float64(A * C)));
        	else
        		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)));
        	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], 2e-31], N[(N[Sqrt[N[(2.0 * N[(N[(A + A), $MachinePrecision] * N[(F * N[(-4.0 * N[(A * C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[(4.0 * N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $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 2 \cdot 10^{-31}:\\
        \;\;\;\;\frac{\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(F \cdot \mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)}\\
        
        \mathbf{else}:\\
        \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\
        
        
        \end{array}
        \end{array}
        
        Derivation
        1. Split input into 2 regimes
        2. if (pow.f64 B #s(literal 2 binary64)) < 2e-31

          1. Initial program 14.7%

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

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

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

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

              \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + \color{blue}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            4. lower-+.f6426.1

              \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(A + A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
          5. Applied rewrites26.1%

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

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

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

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

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

              \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
          7. Applied rewrites26.1%

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

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

              \[\leadsto \frac{\sqrt{2 \cdot \left(\left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{\color{blue}{4 \cdot \left(A \cdot C\right)}} \]
            2. lower-*.f6425.3

              \[\leadsto \frac{\sqrt{2 \cdot \left(\left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot F\right) \cdot \left(A + A\right)\right)}}{4 \cdot \color{blue}{\left(A \cdot C\right)}} \]
          10. Applied rewrites25.3%

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

          if 2e-31 < (pow.f64 B #s(literal 2 binary64))

          1. Initial program 16.4%

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

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

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

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

              \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
          5. Applied rewrites20.8%

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

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

              \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
          8. Recombined 2 regimes into one program.
          9. Final simplification26.7%

            \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{2 \cdot \left(\left(A + A\right) \cdot \left(F \cdot \mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)\right)\right)}}{4 \cdot \left(A \cdot C\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
          10. Add Preprocessing

          Alternative 5: 42.2% accurate, 2.8× speedup?

          \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right) \cdot \left(2 \cdot F\right)\right)} \cdot \frac{0.25}{A \cdot C}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
          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) 2e-31)
             (*
              (sqrt (* (+ A A) (* (fma -4.0 (* A C) (* B_m B_m)) (* 2.0 F))))
              (/ 0.25 (* A C)))
             (* (sqrt (/ F (- B_m))) (- (sqrt 2.0)))))
          B_m = fabs(B);
          assert(A < B_m && B_m < C && C < F);
          double code(double A, double B_m, double C, double F) {
          	double tmp;
          	if (pow(B_m, 2.0) <= 2e-31) {
          		tmp = sqrt(((A + A) * (fma(-4.0, (A * C), (B_m * B_m)) * (2.0 * F)))) * (0.25 / (A * C));
          	} else {
          		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
          	}
          	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) <= 2e-31)
          		tmp = Float64(sqrt(Float64(Float64(A + A) * Float64(fma(-4.0, Float64(A * C), Float64(B_m * B_m)) * Float64(2.0 * F)))) * Float64(0.25 / Float64(A * C)));
          	else
          		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)));
          	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], 2e-31], N[(N[Sqrt[N[(N[(A + A), $MachinePrecision] * N[(N[(-4.0 * N[(A * C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision] * N[(2.0 * F), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(0.25 / N[(A * C), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $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 2 \cdot 10^{-31}:\\
          \;\;\;\;\sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right) \cdot \left(2 \cdot F\right)\right)} \cdot \frac{0.25}{A \cdot C}\\
          
          \mathbf{else}:\\
          \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\
          
          
          \end{array}
          \end{array}
          
          Derivation
          1. Split input into 2 regimes
          2. if (pow.f64 B #s(literal 2 binary64)) < 2e-31

            1. Initial program 14.7%

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

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

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

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

                \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + \color{blue}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              4. lower-+.f6426.1

                \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(A + A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
            5. Applied rewrites26.1%

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

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

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

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

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

                \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
            7. Applied rewrites26.1%

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

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

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

                \[\leadsto \color{blue}{\frac{\frac{1}{4}}{A \cdot C}} \cdot \sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot \left(2 \cdot F\right)\right)} \]
              2. lower-*.f6424.6

                \[\leadsto \frac{0.25}{\color{blue}{A \cdot C}} \cdot \sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot \left(2 \cdot F\right)\right)} \]
            11. Applied rewrites24.6%

              \[\leadsto \color{blue}{\frac{0.25}{A \cdot C}} \cdot \sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot \left(2 \cdot F\right)\right)} \]

            if 2e-31 < (pow.f64 B #s(literal 2 binary64))

            1. Initial program 16.4%

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

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

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

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

                \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
            5. Applied rewrites20.8%

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

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

                \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
            8. Recombined 2 regimes into one program.
            9. Final simplification26.4%

              \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\sqrt{\left(A + A\right) \cdot \left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right) \cdot \left(2 \cdot F\right)\right)} \cdot \frac{0.25}{A \cdot C}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
            10. Add Preprocessing

            Alternative 6: 41.5% accurate, 2.9× speedup?

            \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} \mathbf{if}\;{B\_m}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \end{array} \]
            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) 2e-31)
               (/
                (sqrt (* (* A -8.0) (* C (* F (+ A A)))))
                (- (fma -4.0 (* A C) (* B_m B_m))))
               (* (sqrt (/ F (- B_m))) (- (sqrt 2.0)))))
            B_m = fabs(B);
            assert(A < B_m && B_m < C && C < F);
            double code(double A, double B_m, double C, double F) {
            	double tmp;
            	if (pow(B_m, 2.0) <= 2e-31) {
            		tmp = sqrt(((A * -8.0) * (C * (F * (A + A))))) / -fma(-4.0, (A * C), (B_m * B_m));
            	} else {
            		tmp = sqrt((F / -B_m)) * -sqrt(2.0);
            	}
            	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) <= 2e-31)
            		tmp = Float64(sqrt(Float64(Float64(A * -8.0) * Float64(C * Float64(F * Float64(A + A))))) / Float64(-fma(-4.0, Float64(A * C), Float64(B_m * B_m))));
            	else
            		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)));
            	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], 2e-31], N[(N[Sqrt[N[(N[(A * -8.0), $MachinePrecision] * N[(C * N[(F * N[(A + A), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / (-N[(-4.0 * N[(A * C), $MachinePrecision] + N[(B$95$m * B$95$m), $MachinePrecision]), $MachinePrecision])), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $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 2 \cdot 10^{-31}:\\
            \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B\_m \cdot B\_m\right)}\\
            
            \mathbf{else}:\\
            \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)\\
            
            
            \end{array}
            \end{array}
            
            Derivation
            1. Split input into 2 regimes
            2. if (pow.f64 B #s(literal 2 binary64)) < 2e-31

              1. Initial program 14.7%

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

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

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

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

                  \[\leadsto \frac{\mathsf{neg}\left(\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + \color{blue}{A}\right)}\right)}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
                4. lower-+.f6426.1

                  \[\leadsto \frac{-\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \color{blue}{\left(A + A\right)}}}{{B}^{2} - \left(4 \cdot A\right) \cdot C} \]
              5. Applied rewrites26.1%

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

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

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

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

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

                  \[\leadsto \color{blue}{\frac{\sqrt{\left(2 \cdot \left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right) \cdot F\right)\right) \cdot \left(A + A\right)}}{\mathsf{neg}\left(\left({B}^{2} - \left(4 \cdot A\right) \cdot C\right)\right)}} \]
              7. Applied rewrites26.1%

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

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

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

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

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

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

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

                  \[\leadsto \frac{\sqrt{-2 \cdot \left(\left(B \cdot \color{blue}{\left(B \cdot B\right)}\right) \cdot F\right)}}{\mathsf{neg}\left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)\right)} \]
                7. lower-*.f643.6

                  \[\leadsto \frac{\sqrt{-2 \cdot \left(\left(B \cdot \color{blue}{\left(B \cdot B\right)}\right) \cdot F\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)} \]
              10. Applied rewrites3.6%

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

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

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

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

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

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

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

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

                  \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A + \left(\mathsf{neg}\left(\color{blue}{\left(\mathsf{neg}\left(A\right)\right)}\right)\right)\right)\right)\right)}}{\mathsf{neg}\left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)\right)} \]
                8. remove-double-negN/A

                  \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \left(A + \color{blue}{A}\right)\right)\right)}}{\mathsf{neg}\left(\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)\right)} \]
                9. lower-+.f6424.7

                  \[\leadsto \frac{\sqrt{\left(-8 \cdot A\right) \cdot \left(C \cdot \left(F \cdot \color{blue}{\left(A + A\right)}\right)\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)} \]
              13. Applied rewrites24.7%

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

              if 2e-31 < (pow.f64 B #s(literal 2 binary64))

              1. Initial program 16.4%

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

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

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

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

                  \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
              5. Applied rewrites20.8%

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

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

                  \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
              8. Recombined 2 regimes into one program.
              9. Final simplification26.4%

                \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 2 \cdot 10^{-31}:\\ \;\;\;\;\frac{\sqrt{\left(A \cdot -8\right) \cdot \left(C \cdot \left(F \cdot \left(A + A\right)\right)\right)}}{-\mathsf{fma}\left(-4, A \cdot C, B \cdot B\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
              10. Add Preprocessing

              Alternative 7: 40.4% accurate, 3.3× speedup?

              \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \begin{array}{l} t_0 := -\sqrt{2}\\ \mathbf{if}\;{B\_m}^{2} \leq 10^{-16}:\\ \;\;\;\;t\_0 \cdot \sqrt{\frac{F \cdot -0.5}{C}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot t\_0\\ \end{array} \end{array} \]
              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))))
                 (if (<= (pow B_m 2.0) 1e-16)
                   (* t_0 (sqrt (/ (* F -0.5) C)))
                   (* (sqrt (/ F (- B_m))) t_0))))
              B_m = fabs(B);
              assert(A < B_m && B_m < C && C < F);
              double code(double A, double B_m, double C, double F) {
              	double t_0 = -sqrt(2.0);
              	double tmp;
              	if (pow(B_m, 2.0) <= 1e-16) {
              		tmp = t_0 * sqrt(((F * -0.5) / C));
              	} else {
              		tmp = sqrt((F / -B_m)) * t_0;
              	}
              	return tmp;
              }
              
              B_m = abs(b)
              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
              real(8) function code(a, b_m, c, f)
                  real(8), intent (in) :: a
                  real(8), intent (in) :: b_m
                  real(8), intent (in) :: c
                  real(8), intent (in) :: f
                  real(8) :: t_0
                  real(8) :: tmp
                  t_0 = -sqrt(2.0d0)
                  if ((b_m ** 2.0d0) <= 1d-16) then
                      tmp = t_0 * sqrt(((f * (-0.5d0)) / c))
                  else
                      tmp = sqrt((f / -b_m)) * t_0
                  end if
                  code = tmp
              end function
              
              B_m = Math.abs(B);
              assert A < B_m && B_m < C && C < F;
              public static double code(double A, double B_m, double C, double F) {
              	double t_0 = -Math.sqrt(2.0);
              	double tmp;
              	if (Math.pow(B_m, 2.0) <= 1e-16) {
              		tmp = t_0 * Math.sqrt(((F * -0.5) / C));
              	} else {
              		tmp = Math.sqrt((F / -B_m)) * t_0;
              	}
              	return tmp;
              }
              
              B_m = math.fabs(B)
              [A, B_m, C, F] = sort([A, B_m, C, F])
              def code(A, B_m, C, F):
              	t_0 = -math.sqrt(2.0)
              	tmp = 0
              	if math.pow(B_m, 2.0) <= 1e-16:
              		tmp = t_0 * math.sqrt(((F * -0.5) / C))
              	else:
              		tmp = math.sqrt((F / -B_m)) * t_0
              	return tmp
              
              B_m = abs(B)
              A, B_m, C, F = sort([A, B_m, C, F])
              function code(A, B_m, C, F)
              	t_0 = Float64(-sqrt(2.0))
              	tmp = 0.0
              	if ((B_m ^ 2.0) <= 1e-16)
              		tmp = Float64(t_0 * sqrt(Float64(Float64(F * -0.5) / C)));
              	else
              		tmp = Float64(sqrt(Float64(F / Float64(-B_m))) * t_0);
              	end
              	return tmp
              end
              
              B_m = abs(B);
              A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
              function tmp_2 = code(A, B_m, C, F)
              	t_0 = -sqrt(2.0);
              	tmp = 0.0;
              	if ((B_m ^ 2.0) <= 1e-16)
              		tmp = t_0 * sqrt(((F * -0.5) / C));
              	else
              		tmp = sqrt((F / -B_m)) * t_0;
              	end
              	tmp_2 = tmp;
              end
              
              B_m = N[Abs[B], $MachinePrecision]
              NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
              code[A_, B$95$m_, C_, F_] := Block[{t$95$0 = (-N[Sqrt[2.0], $MachinePrecision])}, If[LessEqual[N[Power[B$95$m, 2.0], $MachinePrecision], 1e-16], N[(t$95$0 * N[Sqrt[N[(N[(F * -0.5), $MachinePrecision] / C), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * t$95$0), $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}\\
              \mathbf{if}\;{B\_m}^{2} \leq 10^{-16}:\\
              \;\;\;\;t\_0 \cdot \sqrt{\frac{F \cdot -0.5}{C}}\\
              
              \mathbf{else}:\\
              \;\;\;\;\sqrt{\frac{F}{-B\_m}} \cdot t\_0\\
              
              
              \end{array}
              \end{array}
              
              Derivation
              1. Split input into 2 regimes
              2. if (pow.f64 B #s(literal 2 binary64)) < 9.9999999999999998e-17

                1. Initial program 15.2%

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

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

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

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

                    \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
                5. Applied rewrites13.7%

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

                  \[\leadsto \sqrt{\frac{-1}{2} \cdot \frac{F}{C}} \cdot \left(\mathsf{neg}\left(\sqrt{\color{blue}{2}}\right)\right) \]
                7. Step-by-step derivation
                  1. Applied rewrites23.0%

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

                  if 9.9999999999999998e-17 < (pow.f64 B #s(literal 2 binary64))

                  1. Initial program 16.0%

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

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

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

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

                      \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
                  5. Applied rewrites20.4%

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

                    \[\leadsto \sqrt{-1 \cdot \frac{F}{B}} \cdot \left(\mathsf{neg}\left(\sqrt{\color{blue}{2}}\right)\right) \]
                  7. Step-by-step derivation
                    1. Applied rewrites28.0%

                      \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
                  8. Recombined 2 regimes into one program.
                  9. Final simplification25.4%

                    \[\leadsto \begin{array}{l} \mathbf{if}\;{B}^{2} \leq 10^{-16}:\\ \;\;\;\;\left(-\sqrt{2}\right) \cdot \sqrt{\frac{F \cdot -0.5}{C}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{F}{-B}} \cdot \left(-\sqrt{2}\right)\\ \end{array} \]
                  10. Add Preprocessing

                  Alternative 8: 27.0% accurate, 12.0× speedup?

                  \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right) \end{array} \]
                  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 (/ F (- B_m))) (- (sqrt 2.0))))
                  B_m = fabs(B);
                  assert(A < B_m && B_m < C && C < F);
                  double code(double A, double B_m, double C, double F) {
                  	return sqrt((F / -B_m)) * -sqrt(2.0);
                  }
                  
                  B_m = abs(b)
                  NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                  real(8) function code(a, b_m, c, f)
                      real(8), intent (in) :: a
                      real(8), intent (in) :: b_m
                      real(8), intent (in) :: c
                      real(8), intent (in) :: f
                      code = sqrt((f / -b_m)) * -sqrt(2.0d0)
                  end function
                  
                  B_m = Math.abs(B);
                  assert A < B_m && B_m < C && C < F;
                  public static double code(double A, double B_m, double C, double F) {
                  	return Math.sqrt((F / -B_m)) * -Math.sqrt(2.0);
                  }
                  
                  B_m = math.fabs(B)
                  [A, B_m, C, F] = sort([A, B_m, C, F])
                  def code(A, B_m, C, F):
                  	return math.sqrt((F / -B_m)) * -math.sqrt(2.0)
                  
                  B_m = abs(B)
                  A, B_m, C, F = sort([A, B_m, C, F])
                  function code(A, B_m, C, F)
                  	return Float64(sqrt(Float64(F / Float64(-B_m))) * Float64(-sqrt(2.0)))
                  end
                  
                  B_m = abs(B);
                  A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                  function tmp = code(A, B_m, C, F)
                  	tmp = sqrt((F / -B_m)) * -sqrt(2.0);
                  end
                  
                  B_m = N[Abs[B], $MachinePrecision]
                  NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                  code[A_, B$95$m_, C_, F_] := N[(N[Sqrt[N[(F / (-B$95$m)), $MachinePrecision]], $MachinePrecision] * (-N[Sqrt[2.0], $MachinePrecision])), $MachinePrecision]
                  
                  \begin{array}{l}
                  B_m = \left|B\right|
                  \\
                  [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\
                  \\
                  \sqrt{\frac{F}{-B\_m}} \cdot \left(-\sqrt{2}\right)
                  \end{array}
                  
                  Derivation
                  1. Initial program 15.6%

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

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

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

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

                      \[\leadsto \color{blue}{\sqrt{\frac{F \cdot \left(\left(A + C\right) - \sqrt{{B}^{2} + {\left(A - C\right)}^{2}}\right)}{{B}^{2} - 4 \cdot \left(A \cdot C\right)}} \cdot \left(\mathsf{neg}\left(\sqrt{2}\right)\right)} \]
                  5. Applied rewrites17.0%

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

                    \[\leadsto \sqrt{-1 \cdot \frac{F}{B}} \cdot \left(\mathsf{neg}\left(\sqrt{\color{blue}{2}}\right)\right) \]
                  7. Step-by-step derivation
                    1. Applied rewrites16.0%

                      \[\leadsto \sqrt{\frac{-F}{B}} \cdot \left(-\sqrt{\color{blue}{2}}\right) \]
                    2. Final simplification16.0%

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

                    Alternative 9: 1.6% accurate, 14.9× speedup?

                    \[\begin{array}{l} B_m = \left|B\right| \\ [A, B_m, C, F] = \mathsf{sort}([A, B_m, C, F])\\ \\ \sqrt{\frac{2}{\frac{B\_m}{F}}} \end{array} \]
                    B_m = (fabs.f64 B)
                    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                    (FPCore (A B_m C F) :precision binary64 (sqrt (/ 2.0 (/ B_m F))))
                    B_m = fabs(B);
                    assert(A < B_m && B_m < C && C < F);
                    double code(double A, double B_m, double C, double F) {
                    	return sqrt((2.0 / (B_m / F)));
                    }
                    
                    B_m = abs(b)
                    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                    real(8) function code(a, b_m, c, f)
                        real(8), intent (in) :: a
                        real(8), intent (in) :: b_m
                        real(8), intent (in) :: c
                        real(8), intent (in) :: f
                        code = sqrt((2.0d0 / (b_m / f)))
                    end function
                    
                    B_m = Math.abs(B);
                    assert A < B_m && B_m < C && C < F;
                    public static double code(double A, double B_m, double C, double F) {
                    	return Math.sqrt((2.0 / (B_m / F)));
                    }
                    
                    B_m = math.fabs(B)
                    [A, B_m, C, F] = sort([A, B_m, C, F])
                    def code(A, B_m, C, F):
                    	return math.sqrt((2.0 / (B_m / F)))
                    
                    B_m = abs(B)
                    A, B_m, C, F = sort([A, B_m, C, F])
                    function code(A, B_m, C, F)
                    	return sqrt(Float64(2.0 / Float64(B_m / F)))
                    end
                    
                    B_m = abs(B);
                    A, B_m, C, F = num2cell(sort([A, B_m, C, F])){:}
                    function tmp = code(A, B_m, C, F)
                    	tmp = sqrt((2.0 / (B_m / F)));
                    end
                    
                    B_m = N[Abs[B], $MachinePrecision]
                    NOTE: A, B_m, C, and F should be sorted in increasing order before calling this function.
                    code[A_, B$95$m_, C_, F_] := N[Sqrt[N[(2.0 / N[(B$95$m / F), $MachinePrecision]), $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}{\frac{B\_m}{F}}}
                    \end{array}
                    
                    Derivation
                    1. Initial program 15.6%

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

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

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

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

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

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

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

                        \[\leadsto \mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \left(\color{blue}{\left(\sqrt{-1} \cdot \sqrt{-1}\right)} \cdot \sqrt{2}\right)\right) \]
                      7. rem-square-sqrtN/A

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

                        \[\leadsto \mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \color{blue}{\left(-1 \cdot \sqrt{2}\right)}\right) \]
                      9. lower-sqrt.f641.7

                        \[\leadsto -\sqrt{\frac{F}{B}} \cdot \left(-1 \cdot \color{blue}{\sqrt{2}}\right) \]
                    5. Applied rewrites1.7%

                      \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \left(-1 \cdot \sqrt{2}\right)} \]
                    6. Step-by-step derivation
                      1. Applied rewrites1.7%

                        \[\leadsto \sqrt{\frac{F}{B} \cdot 2} \]
                      2. Step-by-step derivation
                        1. Applied rewrites1.8%

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

                        Alternative 10: 1.5% accurate, 18.2× 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 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 15.6%

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

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

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

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

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

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

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

                            \[\leadsto \mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \left(\color{blue}{\left(\sqrt{-1} \cdot \sqrt{-1}\right)} \cdot \sqrt{2}\right)\right) \]
                          7. rem-square-sqrtN/A

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

                            \[\leadsto \mathsf{neg}\left(\sqrt{\frac{F}{B}} \cdot \color{blue}{\left(-1 \cdot \sqrt{2}\right)}\right) \]
                          9. lower-sqrt.f641.7

                            \[\leadsto -\sqrt{\frac{F}{B}} \cdot \left(-1 \cdot \color{blue}{\sqrt{2}}\right) \]
                        5. Applied rewrites1.7%

                          \[\leadsto \color{blue}{-\sqrt{\frac{F}{B}} \cdot \left(-1 \cdot \sqrt{2}\right)} \]
                        6. Step-by-step derivation
                          1. Applied rewrites1.7%

                            \[\leadsto \sqrt{\frac{F}{B} \cdot 2} \]
                          2. Final simplification1.7%

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

                          Reproduce

                          ?
                          herbie shell --seed 2024233 
                          (FPCore (A B C F)
                            :name "ABCF->ab-angle b"
                            :precision binary64
                            (/ (- (sqrt (* (* 2.0 (* (- (pow B 2.0) (* (* 4.0 A) C)) F)) (- (+ A C) (sqrt (+ (pow (- A C) 2.0) (pow B 2.0))))))) (- (pow B 2.0) (* (* 4.0 A) C))))