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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-16t^2+96t-84=0

a = -16; b = 96; c = -84;
Δ = b2-4ac
Δ = 962-4·(-16)·(-84)
Δ = 3840
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:

$\sqrt{\Delta}=\sqrt{3840}=\sqrt{256*15}=\sqrt{256}*\sqrt{15}=16\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{15}}{2*-16}=\frac{-96-16\sqrt{15}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{15}}{2*-16}=\frac{-96+16\sqrt{15}}{-32}
$