Помогите по алгебре (5 x^2+x-1) ^2 - (5x^2+x-1) - 2=0

Пусть 5x^2 + x - 1 = t, тогда

t^2 - t - 2 = 0

D = 1 + 4*2 = 9

t₁ = (1 + 3) / 2 = 2;

t₂ = (1 - 3) / 2 = - 1;

Получим 2 случая

1)

5x^2 + x - 1 = 2

5x^2 + x - 3 = 0

D = 1 + 4*5*3 = 61

x₁ = (- 1 + √61) / 10;

x₂ = (- 1 - √61) / 10

2)

5x^2 + x - 1 = - 1

5x^2 + x = 0

x (5x + 1) = 0

x₃ = 0;

5x + 1 = 0 = = > x₄ = - 1/5 = - 0,2