Решить уравнение (x^2-3x)^2+2(x^2-3x)=0

x²-3x = t
t² + 2t = 0
t(t + 2) = 0
t1 = 0; x²-3x = 0; x(x-3) = 0;  x1 = 0; x2 = 3;
t2 = -2; x² - 3x = -2;  x²-3x + 2 = 0 => (x-1)(x-2) = 0; x3 = 1; x4 = 2

Ответ: x1 = 0; x2 = 3; x3 = 1; x4 = 2

(x²  - 3x)²  + 2(x²  - 3 x )  = 0  ⇔ (x² - 3x) (x² - 3x + 2) = 0 ⇔

⇔x(x - 3)(x² - 2x -x +2) = 0 ⇔ x(x - 3)[x(x-2) - (x-2)] = 0 ⇔

⇔ x(x - 3)(x - 2)(x - 1) = 0 

x ∈ {0, 1, 2, 3}