-16t2+6400=0

Solution for -16t2+6400=0 equation:

-16t^2+6400=0

a = -16; b = 0; c = +6400;
Δ = b2-4ac
Δ = 02-4·(-16)·6400
Δ = 409600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{409600}=640$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-640}{2*-16}=\frac{-640}{-32}
=+20
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+640}{2*-16}=\frac{640}{-32}
=-20
$