Solve the following quadratic equations by factorization:

We have been given

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

$\sqrt{2} x^{2}-4 x+x-2 \sqrt{2}=0$

$\sqrt{2} x(x-2 \sqrt{2})+1(x-2 \sqrt{2})=0$

$(x-2 \sqrt{2})(\sqrt{2} x+1)=0$

Therefore,

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

$x=2 \sqrt{2}$

or,

$\sqrt{2} x+1=0$

$\sqrt{2} x=-1$

$x=\frac{-1}{\sqrt{2}}$

Hence, $x=2 \sqrt{2}$ or $x=\frac{-1}{\sqrt{2}}$.