if t is time, then we are solving for the value of t that makes the expression equals 0 (height of the ground)
0=-5t^2+10t+320
factor out -5
0=-5(t^2-2t-64)
use quadratic formula
for at^2+bt+c=0
[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
in our case, a=1, b=-2, c=-64
[tex]t=\frac{-(-2) \pm \sqrt{(-2)^2-4(1)(-64)}}{2(1)}[/tex]
[tex]t=\frac{2 \pm \sqrt{260}}{2}[/tex]
[tex]t=\frac{2 \pm 2\sqrt{65}}{2}[/tex]
[tex]t=1 \pm \sqrt{65}[/tex]
since [tex]1-\sqrt{65}<0[/tex] and we can't have negative time,
[tex]t=1+\sqrt{65}[/tex] which is about 9.062 seconds
