We have, $2x^2$ + x + 1 = 0 Comparing the given equation with the general form $ax^2$ + bx + c = 0, we get a = 2, b = 1, c = 1. ∴ x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ = $\frac{-1\pm\sqrt{(1)^2-4\times2\times1}}{2\times2}$ = $\frac{-1\pm\sqrt{1-8}}{4}$ = $\frac{-1\pm\sqrt{-7}}{4}$ = $\frac{-1\pm\sqrt{7}i}{4}$ ∴ Roots of the given equation are $\frac{-1+\sqrt{7}i}{4}$ and $\frac{-1-\sqrt{7}i}{4}$