Solving a special Quartic Equation. 
Solve for $x$
  $$(x^2-4)(x^2-2x)=2$$  

I have tried the Rational Root Theorem and found that there are no rational roots. Further, the polynomial $p(x)=(x^2-4)(x^2-2x)-2$ is irreducible since when I tried expanding it and writing it as a product of two quadratics, there were no integer solutions for the coefficients. I also depressed the quartic polynomial $p(x)$ hoping that the coefficient of $x$ would also vanish along with the coefficient of $x^3$, giving me a biquadratic. But that didn't happen. I also tried using substitutions,  but none of them worked so far.   
Any help will be appreciated.
Thanks.

$$(x^2-4)(x^2-2x)=2$$
$$\Rightarrow x^4-2x^3-4x^2+8x-2=0$$
$$\Rightarrow (x^2-x-1)^2-3(x-1)^2=0$$
$$\Rightarrow (x^2-x-1+\sqrt 3\ (x-1))(x^2-x-1-\sqrt 3\ (x-1))=0$$
$$\Rightarrow x^2+(\sqrt 3-1)x-1-\sqrt 3=0\ \ \text{or}\ \ x^2-(\sqrt 3+1)x-1+\sqrt 3=0$$
$$\Rightarrow x=\frac{-\sqrt 3+1\pm\sqrt{8+2\sqrt 3}}{2},\frac{\sqrt 3+1\pm\sqrt{8-2\sqrt 3}}{2}.$$
