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

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Answer 1 · 2015-08-17T16:34:52

Let's start by dividing everything by $6$

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

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:
$(n-4)=0->n=4or(n+1)=0->n=-1$

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