0=-16t^2+32t+9

Solution for 0=-16t^2+32t+9 equation:

0=-16t^2+32t+9

We move all terms to the left:

0-(-16t^2+32t+9)=0

We add all the numbers together, and all the variables

-(-16t^2+32t+9)=0

We get rid of parentheses

16t^2-32t-9=0

a = 16; b = -32; c = -9;
Δ = b2-4ac
Δ = -322-4·16·(-9)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-40}{2*16}=\frac{-8}{32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+40}{2*16}=\frac{72}{32}
=2+1/4
$