Решите уравнение (х+2)^4+(х+2)^2-12=0 . за ранее

( x + 2 )^4 + ( x + 2 )^2 - 12 = 0 
( x + 2 )^2 = a ; a > 0 
a^2 + a - 12 = 0 
D = 1 + 48 = 49 = 7^2 
a1 = ( - 1 + 7 ) : 2 = 3 
a2 = ( - 1 - 7 ) : 2 = - 4 ( < 0 ) 
( x + 2 )^2 = 3 
x^2 + 4x + 4 - 3 = 0 
x^2 + 4x + 1 = 0 
D = 16 - 4 = 12 = ( 2√3 )^2 
x1 = ( - 4 + 2√3 ) : 2 = - 2 + √ 3 
x2 = ( - 4 - 2√3 ) : 2 = - 2 - √ 3