Find algebraically the set of values of x for which: (2x-3)(x+2) > 3(x-2) ? Help

find algebraically the set of values of x for which:
(2x-3)(x+2) > 3(x-2)

How do I do this question? Please show your working so that I can understand.
I tried multiplying out and then factorising but I got a surd answer for x and it did not work. The answer should be x < 0, x> 1 but how do I get this? Thank you

(2x-3)(x+2) > 3(x-2)
= 2x^2 +4x-3x-6 > 3x-6
= 2x^2 -2x > 0
= x^2 -x > 0
= x(x-1) > 0
= x < 0 and x > 1

(2x-3)(x+2) > 3(x-2)

2x^2 - 3x + 4x - 6 > 3x - 6

2x^2 + x - 6 - 3x + 6 > 0

2x^2 - 2x > 0

2x(x-1) > 0

2x > 0 and (x-1) > 0

2x/2 > 0/2 and x - 1 > 0

x > 0 and x > 1

x <0 and x >1
Solve (2x-3)(x+2) = 3(x-2) to find the x axis crossings
Plot (2x-3)(x+2) - 3(x-2) with respect to x and see where it is > 0