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. 

Solving algebraically is perhaps not too messy if one is happy to have roots in terms of the solution to a cubic:-
Using the substitution $t=x-1$ we have $$ t^{4}- 4t^{2}-4t+3=0$$ and therefore $$ (t^{2}- 2)^2=4t+1.$$
For any $z$, $$ (t^{2}- 2+2z^2)^2=4z^2t^2+4t+4z^4-8z^2+1.$$
Let $z$ be a solution (found by Cardan's method) of the cubic in $z^2$
$$4z^6-8z^4+z^2-1=0$$
Then $$ (t^{2}- 2+2z^2)^2=(2zt+\frac{1}{z})^2.$$
This is a quadratic in $t$ giving, for example, $t=z+\sqrt {2-z^2+\frac{1}{z}}$ and $t=z-\sqrt {2-z^2+\frac{1}{z}}$.
(One has to be careful choosing the correct signs for whatever root of the cubic is chosen- the signs I have used give real roots for the real root of the cubic.)
