Find the roots of the given equation: $\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$

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Find the roots of the given equation: $\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$

$\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$
$\Rightarrow 2 x^{2}-3 \sqrt{2} x-4=0 \quad$ (Multiplying both sides by $\sqrt{2}$ )
$\Rightarrow 2 x^{2}-3 \sqrt{2} x=4$
$\Rightarrow(\sqrt{2} x)^{2}-2 \times \sqrt{2} x \times \frac{3}{2}+\left(\frac{3}{2}\right)^{2}=4+\left(\frac{3}{2}\right)^{2} \quad$ [Adding $\left(\frac{3}{2}\right)^{2}$ on both sides]
$\Rightarrow\left(\sqrt{2} x-\frac{3}{2}\right)^{2}=4+\frac{9}{4}=\frac{25}{4}=\left(\frac{5}{2}\right)^{2}$
$\Rightarrow \sqrt{2} x-\frac{3}{2}=\pm \frac{5}{2} \quad$ (Taking square root on both sides)
$\Rightarrow \sqrt{2} x-\frac{3}{2}=\frac{5}{2}$ or $\sqrt{2} x-\frac{3}{2}=-\frac{5}{2}$
$\Rightarrow \sqrt{2} x=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4$ or $\sqrt{2} x=-\frac{5}{2}+\frac{3}{2}=-\frac{2}{2}=-1$
$\Rightarrow x=\frac{4}{\sqrt{2}}=2 \sqrt{2}$ or $x=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}$