Решите уравнение х(х^2+4х+4)=3(х+2)

x(x+2)²-3(x+2)=0
(x+2)(x²+2x-3)=0
x+2=0
x=-2
x²+2x-3=0
x1+x2=-2 U x1*x2=-3
x1=-3 U x2=1
 x={-3;-2;1}

x( x^2 + 4x + 4) = 3( x + 2)
x( x + 2)(x + 2) - 3( x + 2) = 0
( x + 2)( x^2 + 2x - 3) = 0
x + 2 = 0
x = - 2
x^2 + 2x - 3 = 0
D = b^2 - 4ac = 4 + 12 = 16 = 4^2
x1 = ( - 2 + 4) / 2 = 1
x2 =( - 2 - 4) / 2 = - 3
Ответ  x1 = - 2, x2 = 1, x3 = - 3