Roots over $\mathbb{C}$ equation $x^{4} - 4x^{3} + 2x^{2} + 4x + 4=0 $. I need roots over $\mathbb{C}$ equation  $$x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = 0$$
From Fundamental theorem of algebra we have statement that the equation have 4 roots over complex. 
But I prepare special reduction: 
$$ \color{red}{ x^{4} - 4x^{3} + 2x^{2} + 4x + 4} = (x-1)^{4}-4(x-1)^{2} + 7 $$
for substitution $y = (x-1)^{2}$ we have: 
$y^{2} - 4y + 7 = 0 $
$y_{0} = 2+i\sqrt{3}$
$y_1 = 2 - i\sqrt{3}$
and we have
$y_{0}^{1/2} + 1 = x_{0} $,  $-y_{0}^{1/2} + 1 = x_{1}$,
$y_{1}^{1/2} + 1 = x_{2}$,  $-y_{1}^{1/2} +1 = x_{3} $
I am not sure what is good results. 
Please check my solution. 
EDIT:
The LHS is not correct, I modify this equation. We should have $p(x) = x^{4} - 4x^{3} + 2x^{2} + 4x + 4 = (x-1)^{4}-4(x-1)^{2} + 7 $
EDIT2: 
I need show that the $p(x)$ is reducible (or not) over $\mathbb{R}[x]$ for two polynomials of degrees 2.
But I am not sure how show that $\left(x-1-\sqrt{2-i\sqrt{3} }\right) \left(x-1+\sqrt{(2+i\sqrt3}\right)$ is (not) polynomial of degree 2. 

Answer to the revised question
You have correctly obtained $x=1\pm \sqrt{2\pm i\sqrt3}$.
By the standard method find the complex square roots. This gives one form of the full answer as $$x=1\pm \sqrt[4]7(\frac{2+\sqrt7\pm i\sqrt3}{\sqrt{14+4\sqrt7}}).$$
