H(d)=-4.9d^2+0.5d

Solution for H(d)=-4.9d^2+0.5d equation:

(H)=-4.9H^2+0.5H

We move all terms to the left:

(H)-(-4.9H^2+0.5H)=0

We get rid of parentheses

4.9H^2-0.5H+H=0

We add all the numbers together, and all the variables

4.9H^2+0.5H=0

a = 4.9; b = 0.5; c = 0;
Δ = b2-4ac
Δ = 0.52-4·4.9·0
Δ = 0.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{0.25}}{2*4.9}=\frac{-0.5-\sqrt{0.25}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{0.25}}{2*4.9}=\frac{-0.5+\sqrt{0.25}}{9.8}
$