-16t^2+800=0

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

-16t^2+800=0

a = -16; b = 0; c = +800;
Δ = b2-4ac
Δ = 02-4·(-16)·800
Δ = 51200
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:

$\sqrt{\Delta}=\sqrt{51200}=\sqrt{25600*2}=\sqrt{25600}*\sqrt{2}=160\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160\sqrt{2}}{2*-16}=\frac{0-160\sqrt{2}}{-32}
=-\frac{160\sqrt{2}}{-32}
=-\frac{5\sqrt{2}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{2}}{2*-16}=\frac{0+160\sqrt{2}}{-32}
=\frac{160\sqrt{2}}{-32}
=\frac{5\sqrt{2}}{-1}
$