-16t^2+216t+108=0

Solution for -16t^2+216t+108=0 equation:

-16t^2+216t+108=0

a = -16; b = 216; c = +108;
Δ = b2-4ac
Δ = 2162-4·(-16)·108
Δ = 53568
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:

$\sqrt{\Delta}=\sqrt{53568}=\sqrt{576*93}=\sqrt{576}*\sqrt{93}=24\sqrt{93}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(216)-24\sqrt{93}}{2*-16}=\frac{-216-24\sqrt{93}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(216)+24\sqrt{93}}{2*-16}=\frac{-216+24\sqrt{93}}{-32}
$