-16t^2+96t+20=0

Solution for -16t^2+96t+20=0 equation:

-16t^2+96t+20=0

a = -16; b = 96; c = +20;
Δ = b2-4ac
Δ = 962-4·(-16)·20
Δ = 10496
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:

$\sqrt{\Delta}=\sqrt{10496}=\sqrt{256*41}=\sqrt{256}*\sqrt{41}=16\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-16\sqrt{41}}{2*-16}=\frac{-96-16\sqrt{41}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+16\sqrt{41}}{2*-16}=\frac{-96+16\sqrt{41}}{-32}
$