How do I solve quadratic equations when the coefficients are complex and real? I needed to solve this: $$x^2 + (2i-3)x + 2-4i = 0 $$
I tried the quadratic formula but it didn't work. So how do I solve this without "guessing" roots? If I guess $x=2$ it works; then I can divide the polynomial and find the other root; but I can't "guess" a root.
$b^2-4ac=4i-3$, now I have to work with $\sqrt{4-3i}$ which I don't know how. Apparently $4i-3$ is equal to $(1+2i)^2$, but I don't know how to get to this answer, so I am stuck.

The square root is not a well defined function on complex numbers. If you want to find out the possible values, the easiest way is probably to go with "Polar form", that is, converting your number into the form 
$$r(\cos(\theta) + i \sin(\theta))$$ and then taking root of it, 
where $r$ is the modulus of the complex number and $\theta$ is the angle with positive direction of $x$-axis or you could find it using $|\frac{y}{x}|$
For example: $$\sqrt3+i$$for this $$r=\sqrt{\sqrt3^2+1^2}$$ and $$\tan\theta=\frac{1}{\sqrt3}$$ and $$\theta = \frac{\pi}{6}$$ and the polar form is $$2(\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}))$$ then find the square root of . which will be $$\pm\left[\sqrt2(\cos(\frac{\pi}{12}+\sin(\frac{\pi}{12}))\right]$$ $$\text{OR}$$  You can use the formula  $$r(\cos(\theta)+ i \sin(\theta))^{1/2} = ±[\sqrt{r}(\cos(\theta/2) + i \sin(\theta/2)].$$
