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 quadratic equation works for real and imaginary coefficients (but maybe without as much geometric intuitiveness when there are imaginary coefficients). You can "check" this by trying to use the quadratic equation and then substituting back in your answers to see if we get $0$. We have $a=1$, $b=2i-3$, $c=2-4i$: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2i+3\pm\sqrt{(2i-3)^2-4(2-4i)}}{2} $$ $$=  \dfrac{-2i+3\pm\sqrt{-4+9-12i-8+16i}}{2}=\dfrac{-2i+3\pm\sqrt{4i-3}}{2}$$
Not pretty, but a quick check in your equation of $x^2+(2i-3)x+2-4i$ with $x=\dfrac{-2i+3+\sqrt{4i-3}}{2}$  gives: $$\left(\dfrac{-2i+3+\sqrt{4i-3}}{2}\right)^2+(2i-3)\cdot\left(\dfrac{-2i+3+\sqrt{4i-3}}{2}\right)+2-4i$$ $$= \dfrac{(-2i+3)^2+4i-3+2(-2i+3)\sqrt{4i-3}}{4}+\dfrac{-(-2i+3)^2+(2i-3)\sqrt{4i-3}}{2}+2-4i$$ $$=\frac{-(-4+9-12i)+4i-3-2(2i-3)\sqrt{10i-3}+2(2i-3)\sqrt{10i-3}}{4}+2-4i$$ $$=\dfrac{-8+16i}{4}+2-4i=0\qquad\checkmark$$
and the other root should (hopefully) follow the same arithmetic, but still work just the same.
