0=16t^2+96t+12

Solution for 0=16t^2+96t+12 equation:

0=16t^2+96t+12

We move all terms to the left:

0-(16t^2+96t+12)=0

We add all the numbers together, and all the variables

-(16t^2+96t+12)=0

We get rid of parentheses

-16t^2-96t-12=0

a = -16; b = -96; c = -12;
Δ = b2-4ac
Δ = -962-4·(-16)·(-12)
Δ = 8448
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:

$\sqrt{\Delta}=\sqrt{8448}=\sqrt{256*33}=\sqrt{256}*\sqrt{33}=16\sqrt{33}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16\sqrt{33}}{2*-16}=\frac{96-16\sqrt{33}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16\sqrt{33}}{2*-16}=\frac{96+16\sqrt{33}}{-32}
$