Average Error: 34.2 → 9.6
Time: 4.6s
Precision: 64
\[\frac{\left(-b_2\right) + \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
↓
\[\begin{array}{l} \mathbf{if}\;b_2 \le -2.7863758169125638 \cdot 10^{138}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \mathbf{elif}\;b_2 \le 1.5867531188831042 \cdot 10^{-76}:\\ \;\;\;\;\frac{\left(-b_2\right) + \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]
\frac{\left(-b_2\right) + \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}
↓
\begin{array}{l}
\mathbf{if}\;b_2 \le -2.7863758169125638 \cdot 10^{138}:\\
\;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\

\mathbf{elif}\;b_2 \le 1.5867531188831042 \cdot 10^{-76}:\\
\;\;\;\;\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}\\

\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 <= -2.7863758169125638e+138)) {
		VAR = ((0.5 * (c / b_2)) - (2.0 * (b_2 / a)));
	} else {
		double VAR_1;
		if ((b_2 <= 1.5867531188831042e-76)) {
			VAR_1 = ((-b_2 + sqrt(((b_2 * b_2) - (a * c)))) / a);
		} else {
			VAR_1 = (-0.5 * (c / b_2));
		}
		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 < -2.7863758169125638e+138

    1. Initial program 58.6

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

    2. Taylor expanded around -inf 2.6

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

    if -2.7863758169125638e+138 < b_2 < 1.5867531188831042e-76

    1. Initial program 12.1

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

    if 1.5867531188831042e-76 < b_2

    1. Initial program 53.8

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

    2. Taylor expanded around inf 8.8

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

  4. Final simplification9.6

    \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -2.7863758169125638 \cdot 10^{138}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \mathbf{elif}\;b_2 \le 1.5867531188831042 \cdot 10^{-76}:\\ \;\;\;\;\frac{\left(-b_2\right) + \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020078 +o rules:numerics
(FPCore (a b_2 c)
  :name "quad2p (problem 3.2.1, positive)"
  :precision binary64
  (/ (+ (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))