Solve $x^4-9x^3+27x^2-29x+6=0$ given that one roots is $2-\sqrt{3}$

Given $x^4-9x^3+27x^2-29x+6=0$

since $2-\sqrt{3}$ is one root then we have 

$2+\sqrt{3}$ is also one root . we have 

$$\begin{gathered} \left(x-\left(2+\sqrt{3}\right)\right)\left(x-\left(2-\sqrt{3}\right)\right)=0 \\ \left(\left(x-2\right)+\sqrt{3}\right)\left(\left(x-2\right)-\sqrt{3}\right)=0 \\ {\left(x-2\right)}^2-{\left(\sqrt{3}\right)}^2=0 \\ {x}^2+4-4x-3=0 \Rightarrow x^2-4x+1=0 \end{gathered}$$

$$\begin{gathered} x^2-5x+6=0; x^2-3x-2x+6=0 \\ \left(x-3\right)\left(x-2\right)=0 x=2.3 \\ The roots are 2, 3, 2+\sqrt{3}, 2-\sqrt{3} \end{gathered}$$