By sketching a graph of this cubic, we can see it has only one real root, x=2.809 approx. The solution is x≥2.809. Here's one way to find it.
Let y=x3-2x2+2x-12. Now put x=0 so y=-12.
Next, x=1: y=-11; x=2: y=-8; x=3: y=27-18+6-12=3. So there is a change of sign between x=2 and x=3. Somewhere in this interval y=0 (curve crosses the x-axis). So we take a value between 2 and 3 for x, say 2.5. y<0 when x=2.5. That means we have a zero between 2.5 and 3, so we take another value between 2.5 and 3, and so on.
A quicker method is to use Newton's (iterative) Method: xn+1=xn-y(xn)/y'(xn), where y'(x)=3x2-4x+2.
Let starting value x0=2, then x1=10/3, x2=662/297, x3=2.812..., x4=2.8089..., x=2.8089030726 approx.
So x≥2.8089 is an approximate solution.