-16t^2+96.4t+80=0

Solution for -16t^2+96.4t+80=0 equation:

-16t^2+96.4t+80=0

a = -16; b = 96.4; c = +80;
Δ = b2-4ac
Δ = 96.42-4·(-16)·80
Δ = 14412.96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96.4)-\sqrt{14412.96}}{2*-16}=\frac{-96.4-\sqrt{14412.96}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96.4)+\sqrt{14412.96}}{2*-16}=\frac{-96.4+\sqrt{14412.96}}{-32}
$