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

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Answer 1 · 2016-01-04T15:56:30

See explanation for a couple of ways...

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

Hence:

$0 = x^2+2x-120 = (x+12)(x-10)$

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

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

$a^2-b^2 = (a-b)(a+b)$

as follows:

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

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

$=(x+1)^2-121$

$=(x+1)^2-11^2$

$=((x+1)-11)((x+1)+11)$

$=(x-10)(x+12)$

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

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