Let $a$, $b$, and $c$ be complex numbers with $a\neq0$. Show that the solutions of $az^2+bz+c=0$ are $z_1,z_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, just as they are in the case when $a$, $b$, and $c$ are real numbers.
If $b^2-4ac=x+yi \in \mathbb{C}$, then
$$\sqrt{x+yi}=\pm \left( \sqrt{\frac{\sqrt{x^2+y^2}+x}{2}} +\frac{iy}{|y|}\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} \right)$$