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)#.
