-16t^2+(128)T=0

Solution for -16t^2+(128)T=0 equation:

-16t^2+(128)=0

a = -16; b = 0; c = +128;
Δ = b2-4ac
Δ = 02-4·(-16)·128
Δ = 8192
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:

$\sqrt{\Delta}=\sqrt{8192}=\sqrt{4096*2}=\sqrt{4096}*\sqrt{2}=64\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64\sqrt{2}}{2*-16}=\frac{0-64\sqrt{2}}{-32}
=-\frac{64\sqrt{2}}{-32}
=-\frac{2\sqrt{2}}{-1}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64\sqrt{2}}{2*-16}=\frac{0+64\sqrt{2}}{-32}
=\frac{64\sqrt{2}}{-32}
=\frac{2\sqrt{2}}{-1}
$