# How do you solve 2x^2 + 12x= 10  using the quadratic formula?

May 9, 2017

$x = - 3 \pm \sqrt{14}$

#### Explanation:

first rearrange to the form$\text{ } a {x}^{2} + b x + c = 0$

we have

$2 {x}^{2} + 12 x - 10 = 0$

divide out any common factors to make it simpler to use.

$\left(2 {x}^{2} + 12 x - 10 = 0\right) \div 2$

$\implies {x}^{2} + 6 x - 5 = 0$

the formula is

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

$a = 1 , b = 6 , c = - 5$

$x = \frac{- 6 \pm \sqrt{{6}^{2} - 4 \times 1 \times \left(- 5\right)}}{2 \times 1}$

$x = \frac{- 6 \pm \sqrt{36 + 20}}{2}$

$x = \frac{- 6 \pm \sqrt{56}}{2}$

$x = - \frac{6}{2} \pm \frac{\sqrt{56}}{2}$

$x = - \frac{6}{2} \pm 2 \frac{\sqrt{14}}{2}$

$\therefore x = - 3 \pm \sqrt{14}$