Question

(x^2-3x)^2-2(x^2-3x)-8>0

Answer 1

 $(x^2-3x)^2-2(x^2-3x)-8>0 \ x^2-3x = t \ t^2-2t-8>0 \ D=4+8*4=4+32=36 \ t_{1,2} = frac{2б6}{2} = 4; -2 \ (t-4)(t+2)>0 \ (x^2-3x-4)(x^2-3x+2)>0 \ x^2-3x-4=0 \ D=9+16=25 \ x_{1,2} = frac{3б5}{2} = 4;-1 \ x^2-3x+2=0 \ D=9-8=1 \ x_{3,4} = frac{3б1}{2} = 2; 1 \ (x-1)(x+1)(x-2)(x-4)>0 \ (-infty;-1) cup (1;2) cup (4;+infty)$