# How do you solve 2x^2+4x-30 =0?

Sep 5, 2016

$x = - 5 , 3$

#### Explanation:

We have: $2 {x}^{2} + 4 x - 30 = 0$

$\implies x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(2\right) \left(- 30\right)}}{2 \left(2\right)}$

$\implies x = \frac{- 4 \pm \sqrt{16 + 240}}{4}$

$\implies x = \frac{- 4 \pm \sqrt{256}}{4}$

$\implies x = \frac{- 4 \pm 16}{4}$

$\implies x = - 1 \pm 4$

$\implies x = - 5 , 3$

Therefore, the solutions to the equation are $x = - 5$ and $x = 3$.

Sep 5, 2016

$x = - 5 \text{ or } x = 3$

#### Explanation:

All the terms in the equation are even, so we can divide by 2 immediately to make the numbers smaller.

$2 {x}^{2} + 4 x - 30 = 0$

${x}^{2} + 2 x - 15 = 0$

Try to factor before we use loner methods of the quadratic formula or completing the square.

Find factors of 15 which subtract (because of -15) to give 2.
The signs will be different (because of -15), there must be more positives. (because of +2)

5 and 3 will work! $5 \times 3 = 15 \mathmr{and} 5 - 3 = 2$

$\left(x + 5\right) \left(x - 3\right) = 0$

Putting each factor equal to 0 gives the solutions:

$x = - 5 \text{ or } x = 3$