# How do you solve x^2+2x-3=0 using the quadratic formula?

Mar 8, 2018

$x = 1 , - 3$

#### Explanation:

We have ${x}^{2} + 2 x - 3 = 0$. For any $a {x}^{2} + b x + c = 0$, $x$ is given by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Here, $a = 1 , b = 2 , c = - 3$. Inputting:

$x = \frac{- \left(2\right) \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 3\right)}}{2 \cdot 1}$

$x = \frac{- 2 \pm \sqrt{4 + 12}}{2}$

$x = \frac{- 2 \pm \sqrt{16}}{2}$

$x = \frac{- 2 \pm 4}{2}$

$x = \frac{2}{2} , - \frac{6}{2}$

$x = 1 , - 3$