0=-16t2+16t+32.

Solution for 0=-16t2+16t+32. equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-16t-32.=0

We add all the numbers together, and all the variables

16t^2-16t-32=0

a = 16; b = -16; c = -32;
Δ = b2-4ac
Δ = -162-4·16·(-32)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-48}{2*16}=\frac{-32}{32}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+48}{2*16}=\frac{64}{32}
=2
$