-16t2+500=0

Solution for -16t2+500=0 equation:

-16t^2+500=0

a = -16; b = 0; c = +500;
Δ = b2-4ac
Δ = 02-4·(-16)·500
Δ = 32000
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:

$\sqrt{\Delta}=\sqrt{32000}=\sqrt{6400*5}=\sqrt{6400}*\sqrt{5}=80\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{5}}{2*-16}=\frac{0-80\sqrt{5}}{-32}
=-\frac{80\sqrt{5}}{-32}
=-\frac{5\sqrt{5}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{5}}{2*-16}=\frac{0+80\sqrt{5}}{-32}
=\frac{80\sqrt{5}}{-32}
=\frac{5\sqrt{5}}{-2}
$