-16t^2+320=0

Solution for -16t^2+320=0 equation:

-16t^2+320=0

a = -16; b = 0; c = +320;
Δ = b2-4ac
Δ = 02-4·(-16)·320
Δ = 20480
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:

$\sqrt{\Delta}=\sqrt{20480}=\sqrt{4096*5}=\sqrt{4096}*\sqrt{5}=64\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{5}}{2*-16}=\frac{0-64\sqrt{5}}{-32}
=-\frac{64\sqrt{5}}{-32}
=-\frac{2\sqrt{5}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{5}}{2*-16}=\frac{0+64\sqrt{5}}{-32}
=\frac{64\sqrt{5}}{-32}
=\frac{2\sqrt{5}}{-1}
$