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 formula is perfectly valid here for the same reason for which it is valid when working only with real numbers:
$$
\frac{-b\pm\sqrt{b^2 - 4ac}} 2 = \frac{-(2i-3) \pm\sqrt{-3+4i}} 2.
$$
The question now is how to find $\pm\sqrt{-3+4i}.$
In polar form, you have $-3+4i = \sqrt{3^3+4^2} \cdot(\cos\alpha+i\sin\alpha) = 5(\cos\alpha+i\sin\alpha)$ where $\cos\alpha = -3/5$ and $\sin\alpha = 4/5,$ so $\tan\alpha = -4/3.$ We then have
$$
\sqrt{5(\cos\alpha+i\sin\alpha)} = \sqrt 5 \cdot\left( \cos\frac\alpha 2 + i\sin\frac\alpha 2 \right).
$$
Now recall from trigonometry that $\tan\dfrac\alpha 2 = \dfrac{\sin\alpha}{1+\cos\alpha} = \dfrac{4/5}{1+(-3/5)} = 2.$
Since $\tan=\dfrac{\text{opposite}}{\text{adjacent}}$, we want $\tan\dfrac\alpha 2 = \dfrac 2 1.$ Since $\tan={\sin}/{\cos},$ the fact that $\tan=2$ means that $\sin = 2\cos.$ Thus we have
$$
\sqrt 5 \cdot \left( \cos\frac\alpha 2 + i\sin\frac\alpha 2 \right) = \sqrt 5 \left( f + i (2f) \right) 
$$
where $f^2 + (2f)^2 = \cos^2 + \sin^2 = 1,$ so $f=\dfrac 1 {\sqrt 5},$ and we then have
$$
\sqrt 5\left(\cos\frac\alpha 2 + i\sin\frac \alpha 2\right) = \sqrt 5\left( \frac 1 {\sqrt 5} + i \frac 2 {\sqrt 5} \right) = 1 + 2i.
$$
Thus $\pm\sqrt{-3+4i} = \pm(1+2i).$
