Solve each of the following quadratic equations:

We write, $-2 \sqrt{6} x=-\sqrt{6} x-\sqrt{6} x$ as $3 x^{2} \times 2=6 x^{2}=(-\sqrt{6} x) \times(-\sqrt{6} x)$

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

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

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

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

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

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

$\Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

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