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Verified Answer :

Divide by 6 to get:

n^2-3n-3=1

Subtracting 1, we see that:

n^2-3n-4=0

Factoring, we have:

(n-4)(n+1)=0

Therefore, 4 and -1 are the solutions, since if either n-4 or n+1 is zero, the whole expression is zero.

Benjamin16

\(6n^{2}-18n-18=6 \\ \\ 6n^{2}-18n-24=0 \\ \\ 6(n^{2}-3n-4)=0 \\ \\ n^{2}-3n-4=0 \\ \\ n^{2}-4n+n-4=0 \\ \\ n(n-4)+(n-4)=0 \\ \\ (n+1)(n-4)=0 \\ \\ n+1=0 \ \vee \ n-4=0 \\ \\ n=-1 \ \vee \ n=4 \\ \\ n\in \lbrace -1,4 \rbrace\)

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