Решите................4 задание)))

№4
$x^2-2- \frac{x^2+8}{x^2+2} =0$
$(x^2-2)- \frac{x^2+8}{x^2+2} =0$ $|*(x^2+2)$
$(x^2+2)(x^2-2)-(x^2+8)=0$
$x^4-4-x^2-8=0$
$x^4-x^2-12=0$
$y=x^2$
$y^2-y-12=0$
$D=b^2-4ac=(-1)^2-4*1*(-12)=1+48=49$
$y_1= \frac{-b+ \sqrt{D} }{2a} = \frac{1+ \sqrt{49} }{2} = \frac{1+7}{2}=4$
$y_2= \frac{-b- \sqrt{D} }{2a} = \frac{1- \sqrt{49} }{2} =\frac{-6}{2} =-3$
$y_1=x^2_1=-3$
$y_2=x^2_2=4$

$x_2^2=4$
$|x=2$
$|x=-2$
Ответ: $-2$