решите уравнение. (x^2+4x)^2+7x^2+28x+12=0. Created by ArxitektoR. algebra-ru

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

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

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