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?

It looks to me as though you need to multiply functions $(f\cdot g \cdot g),$ given the solutions. Here, you are doing function composition, which given your earlier problem, seems you denote as $f\circ g\circ g$. Composing the functions, you did that correctly, no mistakes (I ran through it myself.) So solution to composing the functions is correct, but matches none of the answers you provide.
$$(f\circ g\circ g)(x) = 162n^2 + 576n + 509$$

So, given the notation used in earlier problems you posted to day, try to multiply $f(x)\cdot g(x)\cdot g(x)$
So, you want to expand the following: 
$$(f\cdot g\cdot g)(x)=(2n^2 - 3)(3n+ 4)^2 = 18n^4 + 48n^3 + 5n^2 - 72n - 48$$
But again, the correct result is not a match to any of the given solutions you post.
