-16t^2+116t+96=0

Solution for -16t^2+116t+96=0 equation:

-16t^2+116t+96=0

a = -16; b = 116; c = +96;
Δ = b2-4ac
Δ = 1162-4·(-16)·96
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(116)-140}{2*-16}=\frac{-256}{-32}
=+8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(116)+140}{2*-16}=\frac{24}{-32}
=-3/4
$