Getting the $x$-intercept of $f(x) = -16x^2 + 80x + 5$ $$f(x) = -16x^2 + 80x + 5$$
I need to find the bigger value of $x$ that makes $f(x) = 0$.
Naturally, I thought to do:
$$0=-16x^2+80x+5$$
and I applied the quadratic formula
$$0=\frac{-80\pm\sqrt{6080}}{-32}$$
but the answer doesn't seem like it would be correct.  Did I do something wrong?

First of all, please remember that the $x$-intercept is where the graph $y = \operatorname{f}(x)$ meets the $x$-axis. If you're not plotting a graph then it doesn't make sense to talk about $x$- and $y$-intercepts.
You're looking for the solutions to the equation $\operatorname{f}(x)=0$.
If $\operatorname{f}(x)=-16x^2+80x+5$ then you need to solve $-16x^2+80x+5=0$. The quadratic formula can be used where $a=-16$, $b=80$ and $c=5$. We have:
$$\begin{array}
.x &=& \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\
&=& \frac{-80 \pm \sqrt{(80)^2-4(-16)(5)}}{2(-16)} \\ \\
&=& \frac{-80\pm\sqrt{6720}}{-32} \\ \\
&=& \tfrac{5}{2} \pm \tfrac{1}{4}\sqrt{105}
\end{array}$$
