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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2+16t-4=0

a = 16; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·16·(-4)
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{2}}{2*16}=\frac{-16-16\sqrt{2}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{2}}{2*16}=\frac{-16+16\sqrt{2}}{32}
$