How to solve $z^2-(1-3i)z-2i-2=0$

Using the quadratic formula for a general quadratic $ax^2+bx+c=0$ which is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=0$$ for $z^2-(1-3i)z-2i-2=0$ $$z=\frac{1-3i\pm \sqrt{-8-6i+8i+8}}{2}$$ $$\implies z=\frac{1-3i\pm \sqrt{2i}}{2}$$

I'd define $z = x + yi$ and substitute:

$$(x+yi)^2 - (1 - 3i)(x+yi) - 2i - 2 = 0$$ $$x^2 + 2xyi - y^2 - x - yi + 3xi - 3y^2 - 2i - 2 = 0$$

This gives you two equations (one for the real part and one for the imaginary part) with two unknowns:

$$x^2 - 4y^2 - x - 2 = 0$$ $$3x - y + 2xy - 2 = 0$$

From here you can try some substitutions to remove the coupling, or see if $z = re^{\theta i}$ is easier. In any case, the main trick is that the real part and imaginary must both be equal on both sides.

Can you take it from here?