Solve the following quadratic equations by factorization:

We have been given

$3 x^{2}-2 \sqrt{6} x+2=0$

$3 x^{2}-\sqrt{6} x-\sqrt{6} x+2=0$

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

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

Therefore,

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

$\sqrt{3} x=\sqrt{2}$

$x=\sqrt{\frac{2}{3}}$

or,

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

$\sqrt{3} x=\sqrt{2}$

$x=\sqrt{\frac{2}{3}}$

Hence, $x=\sqrt{\frac{2}{3}}$ or $x=\sqrt{\frac{2}{3}}$.