Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$? Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$?
solution:
\begin{align*}
fgg(n) &= f(g(g(n)))\\
&= f(g(3n+4)))\\
&= f(3(3n + 4) + 4)\\
&= f(9n + 16)\\
&= 2(9n + 16)^2 - 3\\
&= 2(81n^2 + 288n + 256) - 3 \\
&= 162n^2 + 576n + 509
\end{align*}
Check: Try $n = 1$, $f(g(g(1)) = f(g(7)) = f(25) = 2(25)^2 - 3 = 1247$,
$162(1^2) + 576(1) + 509 = 1247$
The practice test multiple choice answers are...
A. $6n^3+8n^2-9n-12$
B. $n^4+4n^3+4n^2+16n$
C. $-6n^3+8n^2+9n-12$
D. $n^4+4n^3+4n^2+16n$
Which one is it? What am I doing wrong?

Let's check what @amWhy noted by using Maple. It is a fun.
 > f:=n-> 2n^2-3:
 > g:=n-> 3n+4:
 > s:=n->(g@g)(n):
 > simplify((f@s)(n));
                                 162n^2+576n+509

