Question: On a wiki proof site, I've seen these formulas used for the mean and variance:

$E[X] = \frac{rp}{1-p}$ and $Var[X] = \frac{rp}{(1-p)^2}$. Yet on the Penn State site, they have these two different formulas: $E[X] = \frac{r}{p}$ and $Var[X] = \frac{r(1-p)}{p^2}$.

Answer: 
**Two** issues are involved:

* Is the negative binomial the distribution of the waiting time until a specified number of "successes" or until a specified number of "failures"?
* Is it the number of trials needed to get the specified number of "successes" (or, as the case may be, of "failures") or is it the number of trials that **do not** result in that many successes, or failures, before that number of "successes" or "failures" happens?

So we have four possibilities:

* It's the number of trials needed to get $r$ successes. Then we have
$$
\mathbb E(X) = \frac rp, \quad \operatorname{var}(X) = \frac{rp}{(1-p)^2}.
$$
* It's the number of failures before the $r$th success. Then we have
$$
\mathbb E(X) = r\frac{1-p}{p},\quad \operatorname{var}(X) = \frac{rp}{(1-p)^2}.
$$
* It's the number of trials needed to get $r$ failures. Then we have
$$
\mathbb E(X) = \frac r{1-p}, \quad \operatorname{var}(X) = \frac{r(1-p)}{p^2}.
$$
* It's the number of successes before the $r$th failure. Then we have
$$
\mathbb E(X) = \frac{rp}{1-p},\quad \operatorname{var}(X) = \frac{rp}{(1-p)^2}.
$$

One advantage to viewing it as the number of successes before the $r$th failure or as the number of failures before the $r$th success is that then it is an infinitely divisible distribution, and it naturally extends to the case where $r$ is not an integer. In that case it is supported on the set $\{0,2,3,4,\dots\}$. In the other cases it is supported on the set $\{r,r+1,r+2,r+3,\dots\}$.
It's a matter of semantics about the meaning of "success" (typically $p$ in the context of a binomial distribution) and "failure" (typically $q=1-p$). If the meaning is reversed in the context of the negative binomial, you'll get the same result, once you interchange $p$ with $1-p$. 

So, you saw the same thing (assuming you really saw $E[X] = \frac{r}{p}-1$).