Решите уравнение (x+2)⁴+2x²+8x-16=0

(x + 2)^4 + 2(x^2 + 4x - 8) = 0 
(x + 2)^4 + 2(x^2 + 4x + 4 - 12) = 0 
(x + 2)^4 + 2((x + 2)^2 - 12) = 0 
(x + 2)^4 + 2(x + 2)^2 - 24 = 0 
Пусть (x + 2)^2 = t, тогда
t^2 + 2t - 24 = 0 
D = 4 + 4*24 = 100 = 10^2
t1 = ( - 2 + 10)/2 = 8/2 = 4;
t2 = ( - 2 - 10)/2 = - 12/2 = - 6;

1) (x + 2)^2 = 4
(x + 2)^2 - 2^2 = 0
(x + 2 - 2)(x + 2 + 2) = 0
x (x + 4) = 0
x1 = 0;
x2 = - 4

2) (x + 2)^2 = - 6
нет решений

ответ
- 4; 0