-16t2+120t+80=0

Solution for -16t2+120t+80=0 equation:

-16t^2+120t+80=0

a = -16; b = 120; c = +80;
Δ = b2-4ac
Δ = 1202-4·(-16)·80
Δ = 19520
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:

$\sqrt{\Delta}=\sqrt{19520}=\sqrt{64*305}=\sqrt{64}*\sqrt{305}=8\sqrt{305}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(120)-8\sqrt{305}}{2*-16}=\frac{-120-8\sqrt{305}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(120)+8\sqrt{305}}{2*-16}=\frac{-120+8\sqrt{305}}{-32}
$