-16t2+64t+80=0

Solution for -16t2+64t+80=0 equation:

-16t^2+64t+80=0

a = -16; b = 64; c = +80;
Δ = b2-4ac
Δ = 642-4·(-16)·80
Δ = 9216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9216}=96$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-96}{2*-16}=\frac{-160}{-32}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+96}{2*-16}=\frac{32}{-32}
=-1
$