We have , $\sqrt{3}x^2-\sqrt{2}x+3\sqrt{3}$ = 0 Comparing the given equation with the general form $ax^2$ + bx + c = 0, we get a = $\sqrt{3}$ , b = $-\sqrt{2}$ , c = $3\sqrt{3}$ ∴ x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ = $\frac{-(-\sqrt{2})\pm\sqrt{(-\sqrt{2})^2-4\times\sqrt{3}\times3\sqrt{3}}}{2\sqrt{3}}$ = $\frac{\sqrt{2}\pm\sqrt{2-36}}{2\sqrt{3}}$ = $\frac{\sqrt{2}\pm\sqrt{-34}}{2\sqrt{3}}$ = $\frac{\sqrt{2}\pm i\sqrt{34}}{2\sqrt{3}}$ ∴ Roots of the given equation are $\frac{\sqrt{2}+i\sqrt{34}}{2\sqrt{3}}$ and $\frac{\sqrt{2}-i\sqrt{34}}{2\sqrt{3}}$