How do you determine whether/how many nonreal solutions an equation has?

The statement:
x^3 - 4x^2 - 3x + 2 = 0 has two nonreal solutions is false

How do I solve this?

Have you studied the Descarte's Rule of Signs ?

here is a good page for it
http://www.purplemath.com/modules/drofsign.htm
or else Google the title for other sources.

or
A quick sketch will show that it crosses the x-axis 3 times.
So there are 3 real solutions.

or
A quick test for x = ±1 and ±2 shows that x = -1 is a solution
so x+1 will be a factor,
By synthetic division
x^3 - 4x^2 - 3x + 2 = (x+1)(x^2 - 5x + 2)
solving x^2 - 5x + 2 = 0
x = (5 ± √17)/2
so x = -1, (5 ± √17)/2 which shows it as 3 real roots.