$\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$
Given: $\sqrt{3} x^{2}-\sqrt{2} x+3 \sqrt{3}=0$
Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=\sqrt{3}, b=-\sqrt{2}$ and $c=3 \sqrt{3}$.
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:
$\alpha=\frac{\sqrt{2}+\sqrt{2-4 \times \sqrt{3} \times 3 \sqrt{3}}}{2 \sqrt{3}}$ and $\beta=\frac{\sqrt{2}-\sqrt{2-4 \times \sqrt{3} \times 3 \sqrt{3}}}{2 \sqrt{3}}$
$\Rightarrow \alpha=\frac{\sqrt{2}+\sqrt{-34}}{2 \sqrt{3}} \quad$ and $\quad \beta=\frac{\sqrt{2}-\sqrt{-34}}{2 \sqrt{3}}$
$\Rightarrow \alpha=\frac{\sqrt{2}+i \sqrt{34}}{2 \sqrt{3}} \quad$ and $\quad \beta=\frac{\sqrt{2}-i \sqrt{34}}{2 \sqrt{3}}$
Hence, the roots of the equation are $\frac{\sqrt{2} \pm i \sqrt{34}}{2 \sqrt{3}}$.