-16t^2+96t+27=0

Solution for -16t^2+96t+27=0 equation:

-16t^2+96t+27=0

a = -16; b = 96; c = +27;
Δ = b2-4ac
Δ = 962-4·(-16)·27
Δ = 10944
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:

$\sqrt{\Delta}=\sqrt{10944}=\sqrt{576*19}=\sqrt{576}*\sqrt{19}=24\sqrt{19}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-24\sqrt{19}}{2*-16}=\frac{-96-24\sqrt{19}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+24\sqrt{19}}{2*-16}=\frac{-96+24\sqrt{19}}{-32}
$