Average Error: 33.1 → 10.4
Time: 4.9s
Precision: 64
\[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
↓
\[\begin{array}{l} \mathbf{if}\;b_2 \le -2.9189091313663101 \cdot 10^{-83}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 7.0423614716856544 \cdot 10^{82}:\\ \;\;\;\;\left(\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}\right) \cdot \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]
\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}
↓
\begin{array}{l}
\mathbf{if}\;b_2 \le -2.9189091313663101 \cdot 10^{-83}:\\
\;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\

\mathbf{elif}\;b_2 \le 7.0423614716856544 \cdot 10^{82}:\\
\;\;\;\;\left(\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}\right) \cdot \frac{1}{a}\\

\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 ((-b_2 - sqrt(((b_2 * b_2) - (a * c)))) / a);
}
↓
double code(double a, double b_2, double c) {
	double VAR;
	if ((b_2 <= -2.91890913136631e-83)) {
		VAR = (-0.5 * (c / b_2));
	} else {
		double VAR_1;
		if ((b_2 <= 7.042361471685654e+82)) {
			VAR_1 = ((-b_2 - sqrt(((b_2 * b_2) - (a * c)))) * (1.0 / a));
		} else {
			VAR_1 = ((0.5 * (c / b_2)) - (2.0 * (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 < -2.91890913136631e-83

    1. Initial program 51.8

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

    2. Taylor expanded around -inf 10.4

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

    if -2.91890913136631e-83 < b_2 < 7.042361471685654e+82

    1. Initial program 12.7

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

    3. Applied div-inv12.8

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

    if 7.042361471685654e+82 < b_2

    1. Initial program 41.9

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

    2. Taylor expanded around inf 4.3

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

  4. Final simplification10.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -2.9189091313663101 \cdot 10^{-83}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 7.0423614716856544 \cdot 10^{82}:\\ \;\;\;\;\left(\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}\right) \cdot \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2020079 +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))