0=16t^2+32t-16

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

0=16t^2+32t-16

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -16; b = -32; c = +16;
Δ = b2-4ac
Δ = -322-4·(-16)·16
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-32\sqrt{2}}{2*-16}=\frac{32-32\sqrt{2}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+32\sqrt{2}}{2*-16}=\frac{32+32\sqrt{2}}{-32}
$