how would you solve the following math problem: x^2-4x+8=0?

You can't factorize such a problem, so you will have to use this fomula:

-b+/- Root of( b^2- 4ac)/2a
You probably will be familiar with the above.
If you solve this, you will find that you have root of a negative number. The root of negative 1 is i, the root of negative 16, is 4i, and so on.
Upon solving, your final answers will be,

2 + 2i and 2- 2i.

x^2 - 4x + 8 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -4
c = 8

x = [4 ±√(16 - 32)]/2
x = [4 ±√(-16)]/2
x = [4 ±√(4^2 * i^1)]/2
x = [4 ±4i]/2
x = 2 ±2i

∴ x = 2 ±2i

x^2-4x+8=0
a= 1, b= -4, c = 8

D= -(b^2) - 4 . a. c
D= -(-4^2) - 4 . 1 . 8
D= 16 - 32
D = - 16

This has complex root. If you have studied them, x = 2 + 2i or 2 - 2i.