Answer
The solutions are $n=6$ and $n=3$.
Work Step by Step
To factor the polynomial $n^{2}+bn+c$ to $(n+p)$ and $(n+q)$, we need to have $p+q=b$ and $pq=c$.
In this case, $b=-9$ and $c=18$.
$\implies p=-6$ and $q=-3$.
Then, $n^{2}-9n+18=(n-6)(n-3)$
$n^{2}-9n+18=0\implies (n-6)(n-3)=0$
Using zero-product property, we have
$n-6=0$ or $n-3=0$
$\implies n=6$ or $n=3$
The solutions are $n=6$ and $n=3$.
