Solve$x^4-6x^3+11x^2-10x+2=0$, given that $2+\sqrt{3}$ is a root.

Given that one root be $2+\sqrt{3}$  then the other roots is 

$2-\sqrt{3}$

Solve $x^2-2x+2=0$ then we will got remaining two roots

$\begin{array}{l}{x}^2-2x+2=0\\ x=\frac{2\pm \sqrt{4-8}}{2}\\ =\frac{2\pm 2i}{2}\\ =1\pm i\end{array}$

Remaining roots are:$2+\sqrt{3},2-\sqrt{3},1+i,1-i$