Помогите решить уравнение

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{(x²+x+1) = (x+1)(x+1)
{(x²+x+2) = (x+1)(x+2)
(x+1)(x+1)(x+1)(x+2) = 12
(x+1)^2(x+2) = 12
(x+1)^2(x+2) - 12 = 0
(x+1)^2(x+2 - 12) = 0
(x+1)^2(x-10) = 0
(x+1)^2 = 0
x-10 = 0
x+1 = 0
x = -1
x = 10
D = (x+2)^2 - 4(x+1)^2
= x^2 + 4x + 4 - 4(x^2 + 2x + 1)
= -3x^2 - 2x - 8
x = (-b ± √(D)) / 2a
= (-4 ± √(-3x^2 - 2x - 8)) / 2(x+1)
= -4 ± √(3x^2 + 2x + 8) / 2(x+1)
Ответ: -1 и 10

x^2+x+1=t (t>0)
t(t+1)=12
t^2+t-12=0
t=-4 посторонний
t=3
x^2+x+1=3
x^2+x-2=0
x=1
x=-2