Solve the following quadratic equations by factorization:

We have been given

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

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

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

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

Therefore,

$x-1=0$

$x=1$

or

$x-\sqrt{3}=0$

$x=\sqrt{3}$

Hence, $x=1$ or $x=\sqrt{3}$.