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) " alt=" (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) " align="absmiddle" class="latex-formula">