# How do you solve 2x^2 + 7x = 5?

Jan 31, 2017

x=(-7+sqrt89)/4; $x = \frac{- 7 - \sqrt{89}}{4}$

#### Explanation:

Solve $2 {x}^{2} + 7 x = 5$

Subtract $5$ from both sides.

$2 {x}^{2} + 7 x - 5 = 0$

This is a quadratic equation in the form $a {x}^{2} + 7 x - 5$, where $a = 2$, $b = 7$, $c = - 5$.

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

Plug in the values from the quadratic formula.

$x = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \cdot 2 \cdot - 5}}{2 \cdot 2}$

$x = \frac{- 7 \pm \sqrt{49 + 40}}{4}$

$x = \frac{- 7 \pm \sqrt{89}}{4}$

$89$ is a prime number so it cannot be factored further.

x=(-7+sqrt89)/4; $x = \frac{- 7 - \sqrt{89}}{4}$