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)
$$2+\sqrt{2}, 2-\sqrt{2},1+i, 1-i$$

By your hint we obtain that it's $$x^4-6x^3+12x^2-12x+4=0$$ or
$$x^4-2x^3+2x^2-4x^3+8x^2-8x+2x^2-4x+4=0$$ or
$$(x^2-2x+2)(x^2-4x+2)=0,$$ which gives the answer:
$$\{1+i,1-i,2+\sqrt2,2-\sqrt2\}.$$
If we don't know the answer then we can get the following.
For all real $k$ we have:
$$x^4-6x^3+12x^2-12x+4=$$
$$=(x^2-3x+k)^2-9x^2-k^2+6kx-2kx^2+12x^2-12x+4=$$
$$=(x^2-3x+k)-2kx^2+3x^2+6kx-12x-k^2+4=$$
$$=(x^2-3x+k)^2-((2k-3)x^2-(6k-12)x+k^2-4).$$
Now, we need to choose a value of $k$ such that $(2k-3)x^2-(6k-12)x+k^2-4=(ax+b)^2$ and we see that $k=2$ is valid.
Thus, $$x^4-6x^3+12x^2-12x+4=(x^2-3x+2)^2-x^2=(x^2-4x+2)(x^2-2x+2)$$
and the rest you saw. 
The best way it's the substitution $x^2+2=t$.
