Just use the formula for finding solutions to quadratic equations, which are equations of the form $ax^2 + bx+c=0$:
$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
In your case, we have $a=1$, $b=2$, and $c=1$, so we get:
$$x = \frac{-2 \pm \sqrt{2^2-4}}{2}=\frac{-2 \pm \sqrt{0}}{2}=\frac{-2 }{2}=-1$$