$$\sum_{n=1}^{\infty} \frac{12}{(-5)^n}$$
If I rewrite this as $\sum_{n=1}^{\infty} 12(-\frac{1}{5})^n$ and then use the infinite geometric series formula, $\frac{a}{1-r}$ where $a = 12$ and $r = (-\frac{1}{5})$? I get 10 but since it starts at 0, I subtracted one from my total equation to get 9, when my textbook says the answer is -2. There must be something I'm doing wrong or misunderstanding.
For $|r|<1$, we have $$ a+ar+ar^2+\ldots=\frac{a}{1-r}. $$ In your case, $a=-12/5$, $r=-1/5$, giving $$ \frac{-12/5}{1-(-1/5)}=-2. $$
For fun: if $S_n=a+ar+ar^2+\ldots+ar^n$ then $$ rS_n=ar+ar^2+\ldots+ar^{n+1}=S_n-a+ar^{n+1} $$ so that $$ S_n=\frac{a(1-r^{n+1})}{1-r} $$ and if $|r|<1$ we have $$ \lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{a(1-r^{n+1})}{1-r}=\frac{a}{1-r} $$ because $$ \lim_{n\to\infty}r^{n+1}=0. $$