x(x+2)>8 if and only if x^2+2x-8>0 if and only if (x+4)(x-2)>0. There are three cases: x<-4, -4 In the first case x+4<0 and x-2<0, so their product is positive: (x+4)(x-2)>0. Next x+4>0 and x-2<0, so their product is negative: (x+4)(x-2)>0. Finally x+4>0 and x-2>0, so their product is positive: (x+4)(x-2)>0. Hence the solutions are in the first and third cases when x<-4 or 2