How do you solve using the quadratic formula for #4x^2+12x+5=117#?

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Answer 1 · 2015-05-31T12:46:15

First subtract 117 from both sides to get:

#4x^2+12x-112 = 0#

Notice all the terms are divisible by #4#, so divide both sides by #4# to get:

#x^2+3x-28 = 0#

This is in the form #ax^2+bx+c = 0# with #a=1#, #b=3#, #c=-28#

Then

#x = (-b +-sqrt(b^2-4ac))/(2a)#

#=(-3+-sqrt(3^2-(4xx1xx-28)))/(2xx1)#

#=(-3+-sqrt(9+112))/2#

#=(-3+-sqrt(121))/2#

#=(-3+-sqrt(11^2))/2#

#=(-3+-11)/2#

That is #x = -7# or #x = 4#