Average Error: 34.0 → 11.0
Time: 5.7s
Precision: 64
\[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
↓
\[\begin{array}{l} \mathbf{if}\;b_2 \le -1.42599941733932729 \cdot 10^{-48}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 9787957843074626:\\ \;\;\;\;\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]
\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}
↓
\begin{array}{l}
\mathbf{if}\;b_2 \le -1.42599941733932729 \cdot 10^{-48}:\\
\;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\

\mathbf{elif}\;b_2 \le 9787957843074626:\\
\;\;\;\;\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\

\end{array}
double code(double a, double b_2, double c) {
	return ((double) (((double) (((double) -(b_2)) - ((double) sqrt(((double) (((double) (b_2 * b_2)) - ((double) (a * c)))))))) / a));
}
↓
double code(double a, double b_2, double c) {
	double VAR;
	if ((b_2 <= -1.4259994173393273e-48)) {
		VAR = ((double) (-0.5 * ((double) (c / b_2))));
	} else {
		double VAR_1;
		if ((b_2 <= 9787957843074626.0)) {
			VAR_1 = ((double) (1.0 / ((double) (a / ((double) (((double) -(b_2)) - ((double) sqrt(((double) (((double) (b_2 * b_2)) - ((double) (a * c))))))))))));
		} else {
			VAR_1 = ((double) (((double) (0.5 * ((double) (c / b_2)))) - ((double) (2.0 * ((double) (b_2 / a))))));
		}
		VAR = VAR_1;
	}
	return VAR;
}

Error

Bits error versus a

Bits error versus b_2

Bits error versus c

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 3 regimes

  2. if b_2 < -1.4259994173393273e-48

    1. Initial program 54.4

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]

    2. Taylor expanded around -inf 7.6

      \[\leadsto \color{blue}{\frac{-1}{2} \cdot \frac{c}{b_2}}\]

    if -1.4259994173393273e-48 < b_2 < 9787957843074626.0

    1. Initial program 15.9

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
    2. Using strategy rm

    3. Applied clear-num16.0

      \[\leadsto \color{blue}{\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}}\]

    if 9787957843074626.0 < b_2

    1. Initial program 33.3

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]

    2. Taylor expanded around inf 7.8

      \[\leadsto \color{blue}{\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}}\]
  3. Recombined 3 regimes into one program.

  4. Final simplification11.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -1.42599941733932729 \cdot 10^{-48}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 9787957843074626:\\ \;\;\;\;\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2020123 +o rules:numerics
(FPCore (a b_2 c)
  :name "quad2m (problem 3.2.1, negative)"
  :precision binary64
  (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))