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

Oct 15, 2015

$x = \frac{1}{2} , x = - 1$

#### Explanation:

$a {x}^{2} + b x + c = 0$

Where,
a is the constant with x to the power 2.
b is the constant with x to the power 1.
c is the constant.

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

Given Equation:
$2 {x}^{2} + x - 1 = 0$

In this Case:
a = 2
b = 1
c = -1

put these values in the quadratic formula.

$x = \frac{- 1 \pm \sqrt{{\left(1\right)}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \left(2\right)}$
$x = \frac{- 1 \pm \sqrt{1 + 8}}{4}$
$x = \frac{- 1 \pm \sqrt{9}}{4}$
$x = \frac{- 1 \pm 3}{4}$
$x = \frac{- 1 + 3}{4} , \frac{- 1 - 3}{4}$
$x = \frac{2}{4} , x = - \frac{4}{4}$
$x = \frac{1}{2} , x = - 1$