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

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

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

We move all terms to the left:

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

We get rid of parentheses

4.9H^2-20H+H-2=0

We add all the numbers together, and all the variables

4.9H^2-19H-2=0

a = 4.9; b = -19; c = -2;
Δ = b2-4ac
Δ = -192-4·4.9·(-2)
Δ = 400.2
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{-(-19)-\sqrt{400.2}}{2*4.9}=\frac{19-\sqrt{400.2}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{400.2}}{2*4.9}=\frac{19+\sqrt{400.2}}{9.8}
$