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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-32t-2=0

a = 16; b = -32; c = -2;
Δ = b2-4ac
Δ = -322-4·16·(-2)
Δ = 1152
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-24\sqrt{2}}{2*16}=\frac{32-24\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+24\sqrt{2}}{2*16}=\frac{32+24\sqrt{2}}{32}
$