I'll write $(1+x)^{1/12}$ as $u$ to save typing, so your equation would be
$$\begin{align}
960&=84.6\left(\frac{1}{u}+\frac{1}{u^2}+\cdots+\frac{1}{u^{12}}\right)\\
&=84.6\left(1+\frac{1}{u}+\frac{1}{u^2}+\cdots+\frac{1}{u^{12}}\right)-84.6
\end{align}$$
The reason I added and subtracted $84.6$ on the right was that the parenthesized expression is now a familiar geometric series with value
$$
1+\frac{1}{u}+\frac{1}{u^2}+\cdots+\frac{1}{u^{12}}=\frac{1-(1/u)^{13}}{1-(1/u)}=\frac{u-(1/u)^{12}}{u-1}
$$
so your equation is now (after a trifling bit of algebra)
$$
\frac{1044.6}{84.6}=\frac{u-(1/u)^{12}}{u-1}
$$
or
$$
960u^{13}-1044.6u^{12}+84.6=0
$$
This has an obvious solution $u=1$ and for the rest you might have to use approximation methods or, what amounts to the same thing, a computer algebra system.
Update. Mathematica tells me there are three real solutions: $u=-0.775465, 1, 1.00871$, corresponding to $x=-0.952712, 0, 0.1096752$, respectively.
84,60means? And does e.g.60/(1+x)^1/12mean $\frac{60}{(1+x)^{1/12}}$? – Rebecca J. Stones Sep 18 '13 at 22:47