trying the factors of 2, we realize that x = 1 is a root. thus by the factor theorem, (x - 1) is a factor.
so, performing long division of polynomials (or synthetic division), we realize that:
$\displaystyle \left( x^4 - 3x^3 + 2 \right) \div (x - 1) = x^3 - 2x^2 - 2x - 2$
so we have: $\displaystyle x^4 - 3x^3 + 2 = (x - 1) \left( x^3 - 2x^2 - 2x - 2 \right)$
the cubic has no rational solution. the one real solution it has is irrational, so it would be hard to find the root for that one. you can use the quadratic formula, or if you are in calculus. you can guess a real root and then refine it using Newton-Raphson method.
to type square root of something, say x, you type: sqrt(x)
or you can learn
LaTex
EDIT: oh, you're in middle school. well...this is an unfair question
i don't recall being able to solve a cubic like that in middle school