(х+2)^4-4 (х+2)^2-5=0 решите уравнение

(x + 2)^2( (x + 2)^2 - 4 - 5) = 0
(x + 2)^2( x^2 + 4x + 4 - 4 - 5) = 0
(x + 2)^2( x^2 + 4x - 5) = 0
(x+ 2)(x + 2)(x^2 + 4x - 5) =0
x + 2 = 0
x = - 2
x^2 + 4x - 5 = 0
D = b^2 - 4ac = 16 + 20 = 36 = 6^2
x1 = ( - 4 + 6) / 2 = 1
x2 = ( - 4 - 6) / 2 = - 5
ответ: x 1 = 1, x2 = - 5, x3 = - 2