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

 A: 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 $\Delta =b^2-4ac.$
A: Use that your equation can be written as $$(z-(2+i)) (z-(1-3 i))$$
A: Let’s complete the square:
$$(z^2-(3-2i)z+\left(\dfrac{3-2i}{2}\right)^2=\left(\dfrac{3-2i}{2}\right)^2-(5-5i).$$ This gives us $$\left(z-\dfrac{3-2i}{2}\right)^2=\dfrac{-15+8i}{4}.$$ It is not hard to realize that the RHS of the last equation is the square of $$\dfrac{1+4i}{2}.$$ Finally take the square root of both sides (do not forget both+ and -) and solve for the unknown $z.$
A: Hint
You had the correct idea but you did the square root wrong.
$$\sqrt{(3-2i)^2-4(5-5i)}\ne 5$$
in fact,
$$\sqrt{(3-2i)^2-4(5-5i)}=\sqrt{-15+8i}=a+bi$$
with, $a,b \in \Bbb R$, so,
$$(a+bi)^2=-15+8i$$
solve the above equation and get your result.
A: The OP made two errors in the exercise:


*

*As the quadratic formula is $\lambda= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$, the $-(3-2i)$ should be replaced by $(3-2i)$ for $-b$.

*The calculation of $\Delta=b^2-4ac$ is incorrect.


We have
$$\Delta =b^2-4ac=(3-2i)^2-4(1)(5-5i)=(5-12i)-(20+20i)=-15+8i$$
so that
$$\lambda=\frac{3-2i\pm \sqrt{-15+8i}}{2}$$
where we can find the square roots $\pm (a+bi)$ of $-15+8i$ by solving the equation
$$(a+ib)^2=-15+8i$$
We have that $(1+4i)^2=-15+8i$, hence the two square roots are $\pm(1+4i).$ Therefore
$$\lambda=\frac{3-2i + (1+4i)}{2}=2+i$$
or
$$\lambda=\frac{3-2i - (1+4i)}{2}=1-3i$$
