-16t^2+100t+80=0

Solution for -16t^2+100t+80=0 equation:

-16t^2+100t+80=0

a = -16; b = 100; c = +80;
Δ = b2-4ac
Δ = 1002-4·(-16)·80
Δ = 15120
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:

$\sqrt{\Delta}=\sqrt{15120}=\sqrt{144*105}=\sqrt{144}*\sqrt{105}=12\sqrt{105}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-12\sqrt{105}}{2*-16}=\frac{-100-12\sqrt{105}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+12\sqrt{105}}{2*-16}=\frac{-100+12\sqrt{105}}{-32}
$