-16t^2-20t+800=0

Solution for -16t^2-20t+800=0 equation:

-16t^2-20t+800=0

a = -16; b = -20; c = +800;
Δ = b2-4ac
Δ = -202-4·(-16)·800
Δ = 51600
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:

$\sqrt{\Delta}=\sqrt{51600}=\sqrt{400*129}=\sqrt{400}*\sqrt{129}=20\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{129}}{2*-16}=\frac{20-20\sqrt{129}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{129}}{2*-16}=\frac{20+20\sqrt{129}}{-32}
$