<s>[INST] Solving Complex Quadratic equations After working a few exercises on the topic, the questions become progressively harder. In this particular exercise I was asked to solve the equations. However I can't quite seem to break this problem as I did previously and I'll explain why. 

The equation is as follows: $$z^2 + (1-i)z + (-6 + 2i) = 0$$

My thought was to use the general formula to get the value of $z$, however it doesn't seem to be the case, or isn't as straight-forward as previous quadratics I've worked with. Please do see my attempt and correct my mistakes whether it be on this question are any other thing in particular that I'm doing wrong, self-teaching is somewhat harder.
My attempt:
Use the formula: $$x =  \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
In my case, $a = 1, b = (1-i), c = (-6 + 2i)$
$$\therefore  z = -(1-i) \pm \frac{\sqrt{(1-i)^2 - 4(1)(-6+2i)}}{2(1)}$$
Which boils down to: $$z = -1+i \pm{\sqrt{12-4i}}$$
From here on I'm confused as to how to continue. Am I going completely off track?
 [/INST] $$
z^2 + (1-i)z + (-6 + 2i) = 0\\
\begin{align}
\implies z &= \frac{-(1-i) \pm \sqrt{(1-i)^2 - 4(-6 + 2i)}}{2}\\
&= \frac{1}{2}\left( i - 1 \,\pm \sqrt{1+ 2i^2 - i^2 +24 - 8i}\right)\\
&= \frac{1}{2}\left(i-1 \pm 2i\sqrt{2i - 6}\right)\\
&=  \left(\frac{1}{2}\pm\sqrt{2(i - 3)} \right)i - \frac{1}{2}
\end{align}
$$
Always remember that the quadratic equation is just a result of "Completing the Square" for a general quadratic:
$$ az^2 + bz + c = 0\\
\implies z^2 + \frac{b}{a}z + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 +\frac{c}{a} = \left(z + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\cdot\frac{4a}{4a} = 0\\
\implies z + \frac{b}{2a} = \sqrt{\frac{b^2 + 4ac}{4a^2}}\\
\implies z  = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
</s>