How do you solve #6x^2+48x-54=0#?

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Answer 1 · 2016-05-26T09:22:54

#x=1# or #x=-9#

In the equation #6x^2+48x-54=0#, as all monomials are divisible by #6#, first divide by #6#. Hence

#x^2+8x-9=0#

We can now split the middle term in to #9x# and #-x# and equation becomes

#x^2+9x-x-9=0#

or #x(x+9)-1(x+9)=0#

or #(x-1)(x+9)=0#

Hence #x=1# or #x=-9#