0=-16t^2-32t+12

Solution for 0=-16t^2-32t+12 equation:

0=-16t^2-32t+12

We move all terms to the left:

0-(-16t^2-32t+12)=0

We add all the numbers together, and all the variables

-(-16t^2-32t+12)=0

We get rid of parentheses

16t^2+32t-12=0

a = 16; b = 32; c = -12;
Δ = b2-4ac
Δ = 322-4·16·(-12)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-16\sqrt{7}}{2*16}=\frac{-32-16\sqrt{7}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+16\sqrt{7}}{2*16}=\frac{-32+16\sqrt{7}}{32}
$