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How do you solve #6n^2-18n-18=6 # by factoring?

1 Answer

Aug 17, 2015

Let's start by dividing everything by #6#

Explanation:

#->n^2-3n-3=1#

Now subtract #1# on both sides to get an #=0# equation
#->n^2-3n-4=0#

Now we factor, by finding factors of #-4# that add up to #-3#
These are #-4and+1#
#->(n-4)(n+1)=0#

Two solutions:
#(n-4)=0->n=4or(n+1)=0->n=-1#

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