# How do you solve x^2 + 2x - 120 = 0?

Jan 4, 2016

See explanation for a couple of ways...

#### Explanation:

Find two factors of $120$ which differ by $2$. The pair $12$, $10$ works.

Hence:

$0 = {x}^{2} + 2 x - 120 = \left(x + 12\right) \left(x - 10\right)$

So $x = - 12$ or $x = 10$

Alternatively, complete the square then use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

as follows:

$0 = {x}^{2} + 2 x - 120$

$= {x}^{2} + 2 x + 1 - 1 - 120$

$= {\left(x + 1\right)}^{2} - 121$

$= {\left(x + 1\right)}^{2} - {11}^{2}$

$= \left(\left(x + 1\right) - 11\right) \left(\left(x + 1\right) + 11\right)$

$= \left(x - 10\right) \left(x + 12\right)$

Hence $x = 10$ or $x = - 12$