0=-16t^2+32t+4

Solution for 0=-16t^2+32t+4 equation:

0=-16t^2+32t+4

We move all terms to the left:

0-(-16t^2+32t+4)=0

We add all the numbers together, and all the variables

-(-16t^2+32t+4)=0

We get rid of parentheses

16t^2-32t-4=0

a = 16; b = -32; c = -4;
Δ = b2-4ac
Δ = -322-4·16·(-4)
Δ = 1280
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:

$\sqrt{\Delta}=\sqrt{1280}=\sqrt{256*5}=\sqrt{256}*\sqrt{5}=16\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{5}}{2*16}=\frac{32-16\sqrt{5}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{5}}{2*16}=\frac{32+16\sqrt{5}}{32}
$