Question

??

Answer 1

Just to be sure ... Are you trying to solve the following:
\[-16t^2 + 90t = 0\]
???

Answer 2

I'm not sure where that 116 is coming from...

Answer 3

If you use quadratic formula
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-90 \pm \sqrt{(90)^2 - 4(-16)(0)}}{2(-16)} = \frac{-90 \pm \sqrt{(90)^2}}{-32} = \frac{-90 \pm 90}{-32} \]

So:
\[\frac{-90 + 90}{-32} = \frac{0}{-32} = 0\]
and
\[\frac{-90 - 90}{-32} = \frac{-180}{-32} = 5.625\]

Answer 4

sorry, that ran right off the page...
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-90 \pm \sqrt{(90)^2 - 4(-16)(0)}}{2(-16)} = \frac{-90 \pm \sqrt{(90)^2}}{-32}\] \[= \frac{-90 \pm 90}{-32}\]

Answer 5

ok, cool. Still not sure about that 116... hope you know :)