(х²+х)²-8·(х+1)·х+12=0 решить уравнение

(x^2 + x)^2 - 8(x^2 + x) + 12 = 0 

Пусть x^2 + x = t, тогда
t^2 - 8t  +12 = 0 
D = 64 - 48 = 16 = 4^2
t₁ = ( 8 + 4)/2 = 12/2 = 6
t₂ = ( 8 - 4)/2 = 4/2 = 2

Обратно, 2 случая
1) x^2 + x = 6
x^2 + x - 6 = 0 
D = 1 + 4*6 = 25 = 5^2
x₁ = ( - 1 + 5)/2 = 2;
x₂ = ( - 1 - 5)/2 = - 3

2) x^2 + x - 2 = 0 
D = 1 + 4*2 = 9 = 3^2
x₃ = ( - 1 + 3)/2 = 2/2 = 1;
x₄ = ( - 1 - 3)/2 = - 4/2 = - 2

ответ
- 3; - 2; 1; 2