Question

решите уравнение. 
(x^2+4x)^2+7x^2+28x+12=0

Answer 1

$(x^2+4x)^2+7(x^2+4x)+12=0\ x^2+4x=a\ a^2+7a+12=0\ D=1^2\ a=frac{-7-1}{2}=-4\ a=frac{-7+1}{2}=-3\ \ x^2+4x+4=0\ x^2+4x+3=0\\ x^2+4x+4=0\ (x+2)^2=0\ x=-2\\ x^2+4x+3=0\ D=16-4*1*3=2^2\ x=frac{-4+2}{2}=-1\ x=frac{-4-2}{2}=-3$

Answer 2

$( x^{2} +4x) ^{2} +7x ^{2} +28x+12=0 \ ( x^{2} +4x) ^{2} +7( x^{2} +4x)+12=0 \ x^{2} +4x=y \ y ^{2} +7y+12=0 \ D>0 \ y_{1} =-3 \ y _{2} =-4 \ x^{2}+4x+3=0 \ D>0 \ x_{1} =-1 \ x_{2} =-3 \ x^{2} +4x+4=0 \ D=0 \ x_{3,4} =-2$