The given equation is $x^{2}-(\sqrt{3}+1) x+\sqrt{3}=0$.

Comparing it with $a x^{2}+b x+c=0$,
we get $a=1, b=-(\sqrt{3}+1)$ and $c=\sqrt{3}$
$\therefore$ Discriminant,
$D=b^{2}-4 a c=[-(\sqrt{3}+1)]^{2}-4 \times 1 \times \sqrt{3}=3+1+2 \sqrt{3}-4 \sqrt{3}=3-2 \sqrt{3}+1=(\sqrt{3}-1)^{2}>0$
So, the given equation has real roots.
Now, $\sqrt{D}=\sqrt{(\sqrt{3}-1)^{2}}=\sqrt{3}-1$
$\begin{array}{l}
\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-[-(\sqrt{3}+1)]+(\sqrt{3}-1)}{2 \times 1}=\frac{\sqrt{3}+1+\sqrt{3}-1}{2}=\frac{2 \sqrt{3}}{2}=\sqrt{3} \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-[-(\sqrt{3}+1)]-(\sqrt{3}-1)}{2 \times 1}=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}=\frac{2}{2}=1
\end{array}$

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