In my textbook, for negative binomial distrubtion where $T_r$ denote the number of trials until the $r^\text{th}$ success in Bernoulli(p) trials.
$$E(T_r) =\frac{r}{p}$$
$$\operatorname{SD}(T_r)=\frac{\sqrt{rq}}{p} = \frac{\sqrt{r(1-p)}}{p}$$
But on Wikipedia, it says the negative binomial distribution has
Which doesn't equate to the above.

Right at the top of the Wikipedia article it says:
If you take the negative binomial distribution to be the distribution of the number of trials needed to get $r$ successes with probability $p$ of success on each trial, then the expected number of such trials is $r/p.$ The set of possible numbers of trials is then $\{r,r+1,r+2,\ldots\}.$
If you take it to be the number of successes before the $r$th failure with probability $p$ of success on each trial, then the expected number of such successes is $pr/(1-p).$ The set of possible numbers of such successes is then $\{0,1,2,3,\ldots\}.$
One could also define it to be the number of failures before then $r$th success, and then the expected value is $r(1-p)/p,$ and there are several other conventions.
One interesting thing about starting at $0$ rather than at $r$ is that in that case the distribution is actually infinitely divisible and one can allow non-integer values of $r.$