Steps and explanation for solving $z^2-(3-2i)z+(5-5i)=0$? I am reading a book and current I am doing some exercises. The exercise that I am having trouble with is this:

Solve the following equation for $z$:
$z^2-(3-2i)z+(5-5i)=0$

I know that the solution is $2+i$ and $1-3i$ but I do not know the steps to get to this result. I have tried using the polar form of complex number to calculate the roots using the solution formula for quadratic equations. However, what I get is not even near the correct results :/
So if anyone could give a user-friendly walkthrough of how to solve that quadratic polynomial - I would be very very happy!
Oh, and here is my general results, if you manage to read it :P


Consider in $\mathbb C[z]$ the equation $$az^2+bz+c=0.$$ As in $\mathbb R[z]$, the solutions are given by  $$z_{1,2}=\frac{-b\pm\sqrt{\Delta }}{2a},$$
where $\Delta =b^2-4ac.$
