A lot of the text for this is from "How does one correctly interpret the behavior of the heat capacity of a charged black hole?" but this concerns a different question. The Reissner-Nordström black hole solution is: $$ds^2=-(1-\frac{2M}{r}+\frac{Q^2}{r^2})dt^2+(1-\frac{2M}{r}+\frac{Q^2}{r^2})^{-1}dr^2 +r^2d\Omega_{2}^2$$
Let us define $f(r)\equiv (1-\frac{2M}{r}+\frac{Q^2}{r^2})$. Clearly, the solutions to $f(r)=0$ are $r_{\pm}=M\pm \sqrt{M^2-Q^2}$, and these represent the two horizons of the charged black hole. If we are considering a point near $r_+$, we can rewrite $f(r)$ as follows: $$f(r_+)\sim \frac{(r_+ -r_-)(r-r_+)}{r_{+}^2} $$
What I don't understand is how can derive the temperature of the black hole from this relation. The temperature is given by $$T=\frac{r_+-r_-}{4\pi r_+^2}=\frac{1}{2\pi }\frac{\sqrt{M^2-Q^2}}{(M+\sqrt{M^2-Q^2})^2}$$
I couldn't find a reasonable answer as to how we can obtain the temperature from $f(r_+)$. What are the steps and reasoning that are missing when making this jump?