x 2 + x + 1 = 0

 D = 

$b^2–4ac$

D = $1^2–4\times 1\times 1$ = $1-4$ = $-3$

Since D<0, the equation will have two distinct Complex roots. The roots are:

$x$ = $\frac{-b+\surd D}{2a}$ or $\frac{-b–\surd D}{2a}$

$x$= $\frac{-1+\sqrt{-3}}{2\times 1}$ or $\frac{-1–\sqrt{-3}}{2\times 1}$

$x$= $\frac{-1+\sqrt{3}i}{2}$ or $\frac{-1–\sqrt{3}i}{2}$