Solution: –

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

Divided by $\sqrt{2}$for both side of each term we get,

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

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

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

Take out common in each term,

$x+1/\sqrt{2}=0,x-2\sqrt{2}=0$

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

Equate both to zero,

$x+1/\sqrt{2}=0,x-2\sqrt{2}=0$

$x=-1/\sqrt{2},x=2\sqrt{2}$