$$f(x) = -16x^2 + 80x + 5$$
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?
We have the discriminant $\Delta$: $$\Delta = b^2 - 4ac = 80^2-4(-16)5=6400+320=6720$$
$$\sqrt{6720} = 8\sqrt{105}$$
And so our "zeros" are $$x_1, x_2 = \frac{-80\pm\sqrt{6720}}{-32}= \frac{-80 \pm 8\sqrt{105}}{-32} = \frac 52 \pm\frac{\sqrt {105}}{4}$$
The discriminant is $80^2-4(-16)5=6400+320=6720$ and not $6400-320=6080$
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: $$\begin{array} .x &=& \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ &=& \frac{-80 \pm \sqrt{(80)^2-4(-16)(5)}}{2(-16)} \\ \\ &=& \frac{-80\pm\sqrt{6720}}{-32} \\ \\ &=& \tfrac{5}{2} \pm \tfrac{1}{4}\sqrt{105} \end{array}$$
