# How do you solve x^2= -2x + 3 using the quadratic formula?

Feb 27, 2017

$x = - 3 , 1$

#### Explanation:

${x}^{2} = - 2 x + 3$
subtract ${x}^{2}$ from each side so you get 0 on the left
$0 = - {x}^{2} - 2 x + 3$
$x = \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a} \setminus \Rightarrow \frac{- \left(- 2\right) \setminus \pm \setminus \sqrt{{\left(- 2\right)}^{2} - 4 \left(- 1\right) \left(3\right)}}{2 \left(- 1\right)}$
simplify and solve for $x$
$x = \frac{2 \setminus \pm \setminus \sqrt{4 + 12}}{- 2} = \frac{2 \setminus \pm \setminus \sqrt{16}}{- 2} = \frac{2 \setminus \pm 4}{- 2}$
$\frac{2 + 4}{-} 2 \setminus \leftarrow x \setminus \rightarrow \frac{2 - 4}{-} 2$
therefore, $x$ is either of the following...
$x = \frac{6}{-} 2 = - 3$
$x = \frac{- 2}{-} 2 = 1$