Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

The given equation is $3 x^{2}-2 \sqrt{6} x+2=0$.

Comparing it with $a x^{2}+b x+c=0$, we get

$a=3, b=-2 \sqrt{6}$ and $c=2$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-2 \sqrt{6})^{2}-4 \times 3 \times 2=24-24=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

$\therefore \alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{6})+0}{2 \times 3}=\frac{2 \sqrt{6}}{6}=\frac{\sqrt{6}}{3}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-2 \sqrt{6})-0}{2 \times 3}=\frac{2 \sqrt{6}}{6}=\frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.