# How do you factor and solve 2x^2+2x-2=0?

given equation ${x}^{2} + x - 1 = 0$
$D = {b}^{2} - 4 a c$ substituting the values of $b = 1 , a = 1 , c = - 1$ we get
$D = 1 + 4 = 5 > 0$ hence we have real and distinct roots
applying the quadratic formula$\textcolor{red}{\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}} \Rightarrow \left[\frac{- 1 \pm \sqrt{1 + 4}}{2}\right] \Rightarrow \frac{- 1 \pm \sqrt{5}}{2}$
hence the roots are $\frac{- 1 \pm \sqrt{5}}{2}$