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?
 A: 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 $z_2=2-3i$.
A: 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
$$z=\frac{-2i+3\pm(1-4i)}2=\begin{cases}1+i,\\2-3i.\end{cases}$$
A: For a general approach, note that
$$
\begin{array}{l}
 \sqrt z  = \sqrt {x + i\,y}  = \sqrt {\left| z \right|} \;e^{\,i\,\arg (z)/2}  =  \\ 
  =  \pm \left( {\sqrt {\frac{{\left| z \right| + {\mathop{\rm Re}\nolimits} (z)}}{2}}  + \;i\,{\rm sign}^{\rm *} \left( {{\mathop{\rm Im}\nolimits} (z)} \right)\sqrt {\frac{{\left| z \right| - {\mathop{\rm Re}\nolimits} (z)}}{2}} } \right) =  \\ 
  =  \pm \left( {\sqrt {\frac{{\sqrt {x^{\,2}  + y^{\,2} }  + x}}{2}}  + i\,{\rm sign}^{\rm *} \left( y \right)\sqrt {\frac{{\sqrt {x^{\,2}  + y^{\,2} }  - x}}{2}} } \right) \\ 
 \end{array}
$$
where the star on the $sign$ function means a definition modified with respect to the "common one", for which
$sign^ *  \left( 0 \right) = 1$
i.e.:
$$
sign^ *  \left( x \right) = \left\{ {\begin{array}{*{20}c}
   { - 1} & {x < 0}  \\
   1 & {0 \leqslant x}  \\
 \end{array} } \right.
$$
(refer to [Wikipedia article][1])  
The quadratic equation $z^{\,2}  + b\,z + c = 0$ can be solved by the method of completing the square
$$
z^{\,2}  - 2\left( { - \frac{b}
{2}} \right)\,z + \left( {\frac{b}
{2}} \right)^2  = \left( {z - \left( { - \frac{b}
{2}} \right)} \right)^2  = \left( {\frac{b}
{2}} \right)^2  - c
$$
Thus providing
$$
z =  - \frac{b}
{2} \pm {}_{\text{(complex)}}\sqrt {\left( {\frac{b}
{2}} \right)^2  - c} 
$$
