Solve each of the following quadratic equations:

$x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$

$\Rightarrow x^{2}-\sqrt{3} x-x+\sqrt{3}=0$

$\Rightarrow x(x-\sqrt{3})-1(x-\sqrt{3})=0$

$\Rightarrow(x-\sqrt{3})(x-1)=0$

$\Rightarrow x-\sqrt{3}=0$ or $x-1=0$

$\Rightarrow x=\sqrt{3}$ or $x=1$

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.