Question 802317
Given:
(1){{{(2x-2)^2>=(3x-2)*(x-3)}}} or
(2){{{(2x-2)^2-(3x-2)*(x-3)>=0}}}
Now expand all parenthesis in (2) to get
(3){{{(4x^2-8x+4)-(3x^2-2x-9x+6)>=0}}} or
(4){{{x^2+3x-2>=0}}}
The parabola of (4) opens upward but goes negative between the limits
(5){{{(-3-sqrt(17))/2<=x<=(-3+sqrt(17))/2}}}
Since the solution domain is for f(x) to greater or equal to zero the answer is
(6){{{x<=(-3-sqrt(17))/2}}} and {{{x>=(-3+sqrt(17))/2}}}