Need help in figuring out what I am doing wrong when solving for n.. Here is the expression that I am trying to solve for n:
$$ \frac{4}{16+n} = \frac{10}{16+n} \frac{10}{16+n}$$
I am doing the following:
\begin{align}
\frac{4}{16+n} & = \frac{100}{(16 + n)^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{16^2 + 32n + n^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{256 + 32n + n^2} \\[8pt]
\frac{1}{16+n} & = \frac{100}{4(256 + 32n + n^2)} \\[8pt]
\frac{1}{16+n} & = \frac{20}{256 + 32n + n^2} \\[8pt]
16+n & = \frac{256 + 32n + n^2}{20} \\[8pt]
n & = \frac{256 + 32n + n^2 - 320}{20} \\[8pt]
n & = \frac{-64 + 32n + n^2}{20} \\[8pt]
20n & = -64 + 32n + n^2 \\[8pt]
20n -32n - n^2 & = -64 \\[8pt]
12n - n^2 & = -64 \\[8pt]
n(12 - n) & = -64
\end{align}
Not sure what to do next now... Book says that n = 9 
Cannot get 9...
Where am I wrong?
Thank you

There's simple method
$$\frac{4}{16+n} = \frac{10}{16+n} \times \frac{10}{16+n}$$
$n\neq-16$
$${4} = \frac{10\times10}{16+n} $$
$$64+4n = 100 $$
$$36=4n$$
$$n=9$$
Well let's figure out your mstake,
$$\frac{4}{16+n} = \frac{100}{n^2+32n+256} $$
$n\neq-16$
$$n^2+32n+256=25(16+n)$$
$$n^2+7n-144=0$$
$$(n-9)(n+16)$$
since $n\neq-16$
solution is $$n=9$$ 
