500=t+4.9t^2

Solution for 500=t+4.9t^2 equation:

500=t+4.9t^2

We move all terms to the left:

500-(t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-t+500=0

We add all the numbers together, and all the variables

-4.9t^2-1t+500=0

a = -4.9; b = -1; c = +500;
Δ = b2-4ac
Δ = -12-4·(-4.9)·500
Δ = 9801
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9801}=99$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-99}{2*-4.9}=\frac{-98}{-9.8}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+99}{2*-4.9}=\frac{100}{-9.8}
=-10+2/9.8
$