# How do you solve -x^2+x+1=0 using the quadratic formula?

Aug 24, 2017

See a solution process below:

#### Explanation:

The quadratic formula states:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{- 1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{1}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{1}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{{\textcolor{b l u e}{1}}^{2} - \left(4 \cdot \textcolor{red}{- 1} \cdot \textcolor{g r e e n}{1}\right)}}{2 \cdot \textcolor{red}{- 1}}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{1 - \left(- 4\right)}}{- 2}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{1 + 4}}{- 2}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{5}}{- 2}$

$x = \frac{\textcolor{b l u e}{1} \pm \sqrt{5}}{2}$

Or

$\frac{1}{2} \pm \frac{\sqrt{5}}{2}$