(2x^3)-(14x^2)+12x=0

x=?

First, factor out the common 2x.

This leaves you with
2x[x^2 -6x +6) = 0

x=0 is one solution. The others are the two roots of
x^2 -6x +6 = 0,
which are not integers. You can compute them with the quadratic equation

x = [6 +/- sqrt(12)]/2 = 3 +/- sqrt3