\[\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}
\]
↓
\[\begin{array}{l}
t_0 := {c}^{4} \cdot {a}^{4}\\
t_1 := \mathsf{fma}\left(c, a \cdot -4, b \cdot b\right)\\
\mathbf{if}\;\frac{\sqrt{b \cdot b + c \cdot \left(a \cdot -4\right)} - b}{a \cdot 2} \leq -0.205:\\
\;\;\;\;\frac{\frac{b \cdot b - t_1}{\left(-b\right) - \sqrt{t_1}}}{a \cdot 2}\\
\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(-0.25, \frac{4 \cdot t_0 + t_0 \cdot 16}{a \cdot {b}^{7}}, \frac{{c}^{3} \cdot -2}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}}\\
\end{array}
\]
double code(double a, double b, double c) {
return (-b + sqrt(((b * b) - ((4.0 * a) * c)))) / (2.0 * a);
}
↓
double code(double a, double b, double c) {
double t_0 = pow(c, 4.0) * pow(a, 4.0);
double t_1 = fma(c, (a * -4.0), (b * b));
double tmp;
if (((sqrt(((b * b) + (c * (a * -4.0)))) - b) / (a * 2.0)) <= -0.205) {
tmp = (((b * b) - t_1) / (-b - sqrt(t_1))) / (a * 2.0);
} else {
tmp = fma(-0.25, (((4.0 * t_0) + (t_0 * 16.0)) / (a * pow(b, 7.0))), (((pow(c, 3.0) * -2.0) / (pow(b, 5.0) / (a * a))) - (c / b))) - (a * (c / (pow(b, 3.0) / c)));
}
return tmp;
}
function code(a, b, c)
return Float64(Float64(Float64(-b) + sqrt(Float64(Float64(b * b) - Float64(Float64(4.0 * a) * c)))) / Float64(2.0 * a))
end
↓
function code(a, b, c)
t_0 = Float64((c ^ 4.0) * (a ^ 4.0))
t_1 = fma(c, Float64(a * -4.0), Float64(b * b))
tmp = 0.0
if (Float64(Float64(sqrt(Float64(Float64(b * b) + Float64(c * Float64(a * -4.0)))) - b) / Float64(a * 2.0)) <= -0.205)
tmp = Float64(Float64(Float64(Float64(b * b) - t_1) / Float64(Float64(-b) - sqrt(t_1))) / Float64(a * 2.0));
else
tmp = Float64(fma(-0.25, Float64(Float64(Float64(4.0 * t_0) + Float64(t_0 * 16.0)) / Float64(a * (b ^ 7.0))), Float64(Float64(Float64((c ^ 3.0) * -2.0) / Float64((b ^ 5.0) / Float64(a * a))) - Float64(c / b))) - Float64(a * Float64(c / Float64((b ^ 3.0) / c))));
end
return tmp
end
code[a_, b_, c_] := N[(N[((-b) + N[Sqrt[N[(N[(b * b), $MachinePrecision] - N[(N[(4.0 * a), $MachinePrecision] * c), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[(2.0 * a), $MachinePrecision]), $MachinePrecision]
↓
code[a_, b_, c_] := Block[{t$95$0 = N[(N[Power[c, 4.0], $MachinePrecision] * N[Power[a, 4.0], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(c * N[(a * -4.0), $MachinePrecision] + N[(b * b), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(N[(N[Sqrt[N[(N[(b * b), $MachinePrecision] + N[(c * N[(a * -4.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] - b), $MachinePrecision] / N[(a * 2.0), $MachinePrecision]), $MachinePrecision], -0.205], N[(N[(N[(N[(b * b), $MachinePrecision] - t$95$1), $MachinePrecision] / N[((-b) - N[Sqrt[t$95$1], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(a * 2.0), $MachinePrecision]), $MachinePrecision], N[(N[(-0.25 * N[(N[(N[(4.0 * t$95$0), $MachinePrecision] + N[(t$95$0 * 16.0), $MachinePrecision]), $MachinePrecision] / N[(a * N[Power[b, 7.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] + N[(N[(N[(N[Power[c, 3.0], $MachinePrecision] * -2.0), $MachinePrecision] / N[(N[Power[b, 5.0], $MachinePrecision] / N[(a * a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(c / b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] - N[(a * N[(c / N[(N[Power[b, 3.0], $MachinePrecision] / c), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]
\frac{\left(-b\right) + \sqrt{b \cdot b - \left(4 \cdot a\right) \cdot c}}{2 \cdot a}
↓
\begin{array}{l}
t_0 := {c}^{4} \cdot {a}^{4}\\
t_1 := \mathsf{fma}\left(c, a \cdot -4, b \cdot b\right)\\
\mathbf{if}\;\frac{\sqrt{b \cdot b + c \cdot \left(a \cdot -4\right)} - b}{a \cdot 2} \leq -0.205:\\
\;\;\;\;\frac{\frac{b \cdot b - t_1}{\left(-b\right) - \sqrt{t_1}}}{a \cdot 2}\\
\mathbf{else}:\\
\;\;\;\;\mathsf{fma}\left(-0.25, \frac{4 \cdot t_0 + t_0 \cdot 16}{a \cdot {b}^{7}}, \frac{{c}^{3} \cdot -2}{\frac{{b}^{5}}{a \cdot a}} - \frac{c}{b}\right) - a \cdot \frac{c}{\frac{{b}^{3}}{c}}\\
\end{array}