-16t^2+96t-80=0

Solution for -16t^2+96t-80=0 equation:

-16t^2+96t-80=0

a = -16; b = 96; c = -80;
Δ = b2-4ac
Δ = 962-4·(-16)·(-80)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-64}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+64}{2*-16}=\frac{-32}{-32}
=1
$