Solve each of the following quadratic equations:

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

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

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

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

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

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

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

Hence, the roots of the given equation are $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$.