Solve each of the following quadratic equations:

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

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

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

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

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

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

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

Hence, the roots of the given equation are $-3 \sqrt{2}$ and $\sqrt{2}$.