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\(\tt{x=\dfrac{-b}{2a}}\)

\[\frac{ -b }{ 2a}\] \[\frac{ -40 }{ -32 }\]
Time = 8

What is your equation exactly?

\(h(t)= -16t^2 + 40t + 1.5\) ?

To find max height, I need to first obtain the time it takes the ball to reach it, then plug in t.

how did you get time = 8?
\(\tt{\dfrac{-40}{-32}=1.25}\) not 8

maximum height is 26.5
you get the time 1.25 and plug it in for t

Thanks! I got it :) `

(via completing the square)

I never learned how to complete squares :(

Okay, so can I shoot one more at you guys?

don't you learn completing the square in Alg 1?