t=-16t^2+2500t+0

Solution for t=-16t^2+2500t+0 equation:

t=-16t^2+2500t+0

We move all terms to the left:

t-(-16t^2+2500t+0)=0

We get rid of parentheses

16t^2-2500t+t-0=0

We add all the numbers together, and all the variables

16t^2-2499t=0

a = 16; b = -2499; c = 0;
Δ = b2-4ac
Δ = -24992-4·16·0
Δ = 6245001
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6245001}=2499$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2499)-2499}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2499)+2499}{2*16}=\frac{4998}{32}
=156+3/16
$