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#