Average Error: 34.0 → 10.7
Time: 5.4s
Precision: 64
\[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
↓
\[\begin{array}{l} \mathbf{if}\;b_2 \le -1.3486756060694896 \cdot 10^{-130}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 6.012732101308388 \cdot 10^{96}:\\ \;\;\;\;\left(\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}\right) \cdot \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-2 \cdot b_2}{a}\\ \end{array}\]
\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}
↓
\begin{array}{l}
\mathbf{if}\;b_2 \le -1.3486756060694896 \cdot 10^{-130}:\\
\;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\

\mathbf{elif}\;b_2 \le 6.012732101308388 \cdot 10^{96}:\\
\;\;\;\;\left(\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}\right) \cdot \frac{1}{a}\\

\mathbf{else}:\\
\;\;\;\;\frac{-2 \cdot 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 <= -1.3486756060694896e-130)) {
		VAR = ((double) (-0.5 * ((double) (c / b_2))));
	} else {
		double VAR_1;
		if ((b_2 <= 6.012732101308388e+96)) {
			VAR_1 = ((double) (((double) (((double) -(b_2)) - ((double) sqrt(((double) (((double) (b_2 * b_2)) - ((double) (a * c)))))))) * ((double) (1.0 / a))));
		} else {
			VAR_1 = ((double) (((double) (-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 < -1.3486756060694896e-130

    1. Initial program 50.6

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

    2. Taylor expanded around -inf 12.4

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

    if -1.3486756060694896e-130 < b_2 < 6.012732101308388e+96

    1. Initial program 11.4

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

    3. Applied div-inv11.5

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

    if 6.012732101308388e+96 < b_2

    1. Initial program 46.7

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

    3. Applied flip--62.8

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

    4. Simplified61.9

      \[\leadsto \frac{\frac{\color{blue}{0 + a \cdot c}}{\left(-b_2\right) + \sqrt{b_2 \cdot b_2 - a \cdot c}}}{a}\]

    5. Simplified61.9

      \[\leadsto \frac{\frac{0 + a \cdot c}{\color{blue}{\sqrt{b_2 \cdot b_2 - a \cdot c} - b_2}}}{a}\]

    6. Taylor expanded around 0 4.1

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

  4. Final simplification10.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -1.3486756060694896 \cdot 10^{-130}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 6.012732101308388 \cdot 10^{96}:\\ \;\;\;\;\left(\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}\right) \cdot \frac{1}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-2 \cdot b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2020121 
(FPCore (a b_2 c)
  :name "NMSE problem 3.2.1"
  :precision binary64
  (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))