-16t2+32t+80=0

Solution for -16t2+32t+80=0 equation:

-16t^2+32t+80=0

a = -16; b = 32; c = +80;
Δ = b2-4ac
Δ = 322-4·(-16)·80
Δ = 6144
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:

$\sqrt{\Delta}=\sqrt{6144}=\sqrt{1024*6}=\sqrt{1024}*\sqrt{6}=32\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32\sqrt{6}}{2*-16}=\frac{-32-32\sqrt{6}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32\sqrt{6}}{2*-16}=\frac{-32+32\sqrt{6}}{-32}
$