H(t)=20t-4.9t^2

Solution for H(t)=20t-4.9t^2 equation:

(H)=20H-4.9H^2

We move all terms to the left:

(H)-(20H-4.9H^2)=0

We get rid of parentheses

4.9H^2-20H+H=0

We add all the numbers together, and all the variables

4.9H^2-19H=0

a = 4.9; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·4.9·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*4.9}=\frac{0}{9.8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*4.9}=\frac{38}{9.8}
=3+5.375/6.125
$