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?

Here's the general method:
You have to find the roots of $\Delta=-15-8i$. Let $(x+iy)^2=\Delta$. This means
$$x^2-y^2+2ixy=-15-8i\tag{1}$$
This means $x^2-y^2=-15$, $\;xy=-4$. As it would be too complicated to eliminate one of the unknowns to determine the other, we'll obtain another equation from the moduli.
Observe that $\lvert x+iy\rvert^2=\lvert -15+8i\rvert=\sqrt{289}=17$, whence the linear system in $x^2$ and $y ^2$:
$$\begin{cases}x^2-y^2=-15\\x^2+y^2=17\end{cases}\iff\begin{cases}x^2=1\\y^2=16\end{cases}\iff\begin{cases}x=\pm1\\y=\pm4\end{cases}$$
There seems to be $4$ square roots. However, note the equation $xy=-4$ implies $x$ and $y$ have opposite signs. So there are  really only two square roots: $\;\pm(1-4i)$, and the roots of the initial equation are
$$z=\frac{-2i+3\pm(1-4i)}2=\begin{cases}1+i,\\2-3i.\end{cases}$$
