Average Error: 34.3 → 9.7
Time: 6.1s
Precision: 64
\[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
↓
\[\begin{array}{l} \mathbf{if}\;b_2 \le -1.0996442222619662 \cdot 10^{-75}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 1.38731738723855584 \cdot 10^{50}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]
\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}
↓
\begin{array}{l}
\mathbf{if}\;b_2 \le -1.0996442222619662 \cdot 10^{-75}:\\
\;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\

\mathbf{elif}\;b_2 \le 1.38731738723855584 \cdot 10^{50}:\\
\;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{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 ((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 <= -1.0996442222619662e-75)) {
		VAR = ((double) (-0.5 * ((double) (c / b_2))));
	} else {
		double VAR_1;
		if ((b_2 <= 1.3873173872385558e+50)) {
			VAR_1 = ((double) (((double) (((double) -(b_2)) - ((double) sqrt(((double) (((double) (b_2 * b_2)) - ((double) (a * c)))))))) / a));
		} 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 < -1.0996442222619662e-75

    1. Initial program 53.9

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

    2. Taylor expanded around -inf 8.8

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

    if -1.0996442222619662e-75 < b_2 < 1.3873173872385558e+50

    1. Initial program 13.2

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

    3. Applied clear-num13.3

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

    5. Applied clear-num13.3

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

    6. Applied remove-double-div13.2

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

    if 1.3873173872385558e+50 < b_2

    1. Initial program 37.3

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

    2. Taylor expanded around inf 4.8

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

  4. Final simplification9.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -1.0996442222619662 \cdot 10^{-75}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 1.38731738723855584 \cdot 10^{50}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2020114 +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))