How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$ 
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$

I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
$$
z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2} = \cdots = \frac{3-2i \pm \sqrt{-8i-15}}{2}
$$
I am guessing I cannot go further than this.
Other idea: Being $z = ai+b$,
$$
(ai+b)^2 + (2i-3)(ai+b) + 5 - i = 0
$$
Which will give me a system of two equations and two variables
$$
\begin{cases}
-a^2-2a+b^2-3b = -5\\
2ab+2b-3a = 1
\end{cases}
$$
which seems almost impossible to solve.
What am I missing? Is there a simpler solution to this that I am not thinking about?

By solving $(a+ib)^2=8i+15$ we get
$a^2-b^2=15$, $2ab=8$ which are satisfied by $a=4$ and $b=1$.
Hence we can go on with the quadratic formula:
$$z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2}  = \frac{3-2i \pm \sqrt{-8i-15}}{2}\\
= \frac{3-2i \pm i\sqrt{8i+15}}{2}=\frac{3-2i \pm i(4+i)}{2}=\frac{(3\mp 1)+ i(-2\pm 4)}{2}$$
which implies that the solutions are $z_1=1+i$ and $z_2=2-3i$.
