If we factor the numerator of the left hand side completely, we obtain
(3(x+2)(x-2)^2)/(x-7)>0
We wish to identify the intervals on which the left hand side is positive. As the sign can only change at points where the left hand side evaluates to 0 (a factor of the number is 0) or is undefined (a factor of the denominator is 0), we will check the intervals with endpoints at those values.
We can also simplify the task slightly by noting that (x-2)^2>0 for all x!=2, and so that factor will never change the sign of the expression. Proceeding:
On (-oo,-2) we have x+2 < 0 and x-7 < 0, meaning we have a -/(-) =+ case, and so (3(x+2)(x-2)^2)/(x-7)>0.
On (-2,7) we have x+2 > 0 and x-7 < 0, meaning we have a +/(-) = - case, and so (3(x+2)(x-2)^2)/(x-7)<0
(note that the expression is 0 at x=2, and so x=2 would not be in the solution set in any case)
On (7,oo) we have x+2 > 0 and x-7 > 0, meaning we have a +/+ = + case, and so (3(x+2)(x-2)^2)/(x-7)>0.
As we only want the + cases, we get our final solution set as x in (-oo,-2)uu(7,oo).
