When $a
e 0$, there are two solutions to (ax^2 + bx + c = 0) and they are
$$x = {-b pm sqrt{b^2-4ac} over 2a}.$$
egin{equation}
label{eq6}
[x_{i}]=left{
egin{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}
ight.
end{equation}