When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
\begin{equation}
\label{eq6}
[x_{i}]=\left\{
\begin{aligned}
x_{ac} 中选& , & \mu_{a}(x_{i})\geq \mu_{b}(x_{i}), \\
x_{bc} 落选& , & \mu_{a}(x_{i})< \mu_{b}(x_{i}).
\end{aligned}
\right.
\end{equation}
原文：https://www.cnblogs.com/breakdown/p/12535418.html