Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$ Question:

Solve for real $x$ if $(x^2+2)^2+8x^2=6x(x^2+2)$

My attempts:


*

*Here's the expanded form:$$x^4-6x^3+12x^2-12x+4=0$$

*I've plugged this into several online "math problem solving" websites, all claim that "solution could not be determined algebraically, hence numerical methods (i suppose the quartic formula?) were used"

*I've substituted $y=x^2+2$ but then the $6x$  remains to prevent me from solving.

*I've tried factorizing in other ways, I've tried finding simple first solutions but they are actually not simple so I couldn't find them.

*Factoring into circles and hyperbola to arrive at a geometric solution, by estimating the roots from the graph



For reference, the roots are: (credits to wolframalpha)
$$2+\sqrt{2}, 2-\sqrt{2},1+i, 1-i$$

$$(x^2+2)^2+8x^2=6x(x^2+2)$$
$$(x^2+2)^2-6x(x^2+2)+8x^2=0$$
Let $x^2+2=U, x=V$. Then 
$$U^2-6UV+8V^2=0$$
Then
$$\left(\frac UV\right)^2-6\left(\frac UV\right)+8=0$$
Then
$\frac UV=2$ or $\frac UV=4$
$\frac {x^2+2}{x}=2$ or $\frac {x^2+2}{x}=4$
$x^2-2x+2=0$ or $x^2-4x+2=0$
$$x=2\pm \sqrt2$$
