500=(0)(t)+0.5(10)(t^2)

Solution for 500=(0)(t)+0.5(10)(t^2) equation:

500=(0)(t)+0.5(10)(t^2)

We move all terms to the left:

500-((0)(t)+0.5(10)(t^2))=0

We get rid of parentheses

-0.510t^2-0t+500=0

We add all the numbers together, and all the variables

-0.51t^2-1t+500=0

a = -0.51; b = -1; c = +500;
Δ = b2-4ac
Δ = -12-4·(-0.51)·500
Δ = 1021
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{1021}}{2*-0.51}=\frac{1-\sqrt{1021}}{-1.02}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{1021}}{2*-0.51}=\frac{1+\sqrt{1021}}{-1.02}
$