0=-16t^2+8t+32

Solution for 0=-16t^2+8t+32 equation:

0=-16t^2+8t+32

We move all terms to the left:

0-(-16t^2+8t+32)=0

We add all the numbers together, and all the variables

-(-16t^2+8t+32)=0

We get rid of parentheses

16t^2-8t-32=0

a = 16; b = -8; c = -32;
Δ = b2-4ac
Δ = -82-4·16·(-32)
Δ = 2112
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:

$\sqrt{\Delta}=\sqrt{2112}=\sqrt{64*33}=\sqrt{64}*\sqrt{33}=8\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{33}}{2*16}=\frac{8-8\sqrt{33}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{33}}{2*16}=\frac{8+8\sqrt{33}}{32}
$