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.

Sometimes questions have small tricks in them which allow for quick solutions. This question is one of them if the expression is rearranged and factorised. 
$x^2+(2i−3)x+(2−4i) = 0$
Group the real and imaginary components:
$x^2-3x+2+2ix-4i=0$
Now factorise:
$(x-1)(x-2)+2i(x-2)=0$
...and factorise again:
$(x-2)(x-1+2i)=0$
...and applying the null factor theorem, $x=2$ or $x=1-2i.$
I know that this is just a lucky coincidence with this question but it is a technique nonetheless to solve problems. My best advice would be to know as many techniques as possible, such as:


*

*factorisation (on occasion it gives elegant solutions like the one above)

*sum/product of roots (uses simultaneous equations)

*trial and error/factor theorem

*quadratic formula


and more, using simpler techniques wherever you can.

Square roots of complex numbers can either be dealt with algebraically or using the polar forms in other solutions here. Doing so algebraically can be done by using an identity with standard forms of complex numbers ($x+iy$, $a+ib$ etc.) and simultaneous equations.
let $Z = \sqrt{4i-3}$ where Z is complex.
i.e. $Z$ is of the form $Z = a + ib$, where $a$ and $b$ are REAL coefficients. (This is important later on.)
Square both sides: $Z^2 = -3+4i$
Expand the standard form of $Z^2$: $(a+ib)^2 = -3+4i$
This becomes $a^2 - b^2 + 2abi = -3+4i$
The above is an identity and so the real and imaginary parts can be equated.
Hence two equations are formed:
$a^2 - b^2 = -3$................(1)
$2abi = 4i$ which becomes $ab = 2$......(2)
In (2), $b = 2/a$ so $a^2 - (2/a)^2 + 3 = 0$
i.e. $a^4 + 3a^2 - 4 = 0$ (multiplying all by $a^2$ and rearranging)
which is a quadratic-styled quartic: $(a^2 + 4)(a^2 - 1) = 0.$
Solving, $a=1, -1, 2i$ and $-2i$
BUT
since $a$ and $b$ are real, we discard $2i$ and $-2i$.  
Now we simply solve for $b$ using $b = 2/a$. Hence the solutions are $a=1, b=2$ or $a=-1, b=-2$.
and so $\sqrt{4i-3} = \pm(1+2i)$ since we defined $Z = a+ib$.
Best of luck with any future questions! Hope this helped.
