$\sqrt{2} x^{2}+x+\sqrt{2}=0$
Given: $\sqrt{2} x^{2}+x+\sqrt{2}=0$
Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=\sqrt{2}, b=1$ and $c=\sqrt{2}$.
Substituting these values in $\alpha=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$ and $\beta=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$, we get:
$\alpha=\frac{-1+\sqrt{1-4 \times \sqrt{2} \times \sqrt{2}}}{2 \sqrt{2}}$ and $\beta=\frac{-1-\sqrt{1-4 \times \sqrt{2} \times \sqrt{2}}}{2 \sqrt{2}}$
$\Rightarrow \alpha=\frac{-1+\sqrt{-7}}{2 \sqrt{2}} \quad$ and $\quad \beta=\frac{-1-\sqrt{-7}}{2 \sqrt{2}}$
$\Rightarrow \alpha=\frac{-1+i \sqrt{7}}{2 \sqrt{2}} \quad$ and $\quad \beta=\frac{-1-i \sqrt{7}}{2 \sqrt{2}}$
Hence, the roots of the equation are $\frac{-1 \pm i \sqrt{7}}{2 \sqrt{2}}$.