If $n(x)=3 x^{2}+2 x-1$ and $s(x)=x-3$, what is $n(s(x))$ equal to? $3 x^{2}-16 x+26$ $3 x^{2}+2 x+20$ $3 x^{2}-16 x+20$ $3 x^{3}-7 x^{2}-7 x+3$ $3 x^{2}+2 x-4$

Step 1 ：Given the functions $n(x)=3 x^{2}+2 x-1$ and $s(x)=x-3$, we are asked to find the value of $n(s(x))$.

Step 2 ：To find $n(s(x))$, we substitute $s(x)$ into $n(x)$, replacing every instance of $x$ in $n(x)$ with the expression for $s(x)$, which is $x-3$.

Step 3 ：So, $n(s(x)) = 3*(x - 3)^{2} + 2*(x - 3) - 1$.

Step 4 ：Simplifying the expression, we get $n(s(x)) = 3 x^{2} - 16 x + 20$.

Step 5 ：Final Answer: The expression $n(s(x))$ is equal to $3 x^{2}-16 x+20$. So, the correct answer is \(\boxed{3 x^{2}-16 x+20}\).