H(t)=-4.9t^2+32t+0.2

Solution for H(t)=-4.9t^2+32t+0.2 equation:

(H)=-4.9H^2+32H+0.2

We move all terms to the left:

(H)-(-4.9H^2+32H+0.2)=0

We get rid of parentheses

4.9H^2-32H+H-0.2=0

We add all the numbers together, and all the variables

4.9H^2-31H-0.2=0

a = 4.9; b = -31; c = -0.2;
Δ = b2-4ac
Δ = -312-4·4.9·(-0.2)
Δ = 964.92
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{-(-31)-\sqrt{964.92}}{2*4.9}=\frac{31-\sqrt{964.92}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{964.92}}{2*4.9}=\frac{31+\sqrt{964.92}}{9.8}
$