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

Solution for H(t)=-4.9t^2+500 equation:

(H)=-4.9H^2+500

We move all terms to the left:

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

We get rid of parentheses

4.9H^2+H-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:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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