How do you solve #4x^2-13x=12#?

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Answer 1 · 2015-06-27T15:44:40

First subtract #12# from both sides to get #4x^2-13x-12 = 0#, then use the quadratic formula to find:

#x = (13 +- 19)/8#

That is #x=4# or #x=-3/4#

#4x^2-13x-12# is in the form #ax^2+bx+c# with #a=4#, #b=-13# and #c=-12#

It discriminant is given by the formula:

#Delta = b^2-4ac = 13^2 - (4xx4xx-12) = 169+192 = 361 = 19^2#

Since #Delta# is positive and a perfect square, the roots of our quadratic are rational, given by the formula:

#x = (-b+-sqrt(Delta))/(2a) = (13+-19)/8#

That is #x=32/8=4# or #x=-6/8 = -3/4#