$x^{2}-4 x+1=0$
$\Rightarrow x^{2}-4 x=-1$
$\Rightarrow x^{2}-2 \times x \times 2+2^{2}=-1+2^{2} \quad$ (Adding $2^{2}$ on both sides)
$\Rightarrow(x-2)^{2}=-1+4=3$ $\Rightarrow x-2=\pm \sqrt{3} \quad$
$\Rightarrow x-2=\sqrt{3}$ or $x-2=-\sqrt{3}$
$\Rightarrow x=2+\sqrt{3}$ or $x=2-\sqrt{3}$

Hence, $2+\sqrt{3}$ and $2-\sqrt{3}$ are the roots of the given equation.