$x\geq -\frac{\alpha ^{2}}{6}-\frac{2}{3}$

$\left\{\begin{matrix}\alpha \in \mathrm{R}\text{, }&x\geq -\frac{2}{3}\\\alpha \in (-\infty,-\sqrt{-6x-4}]\cup [\sqrt{-6x-4},\infty)\text{, }&x\leq -\frac{2}{3}\end{matrix}\right.$