H(t)=-9.8t^2+4.9t

Solution for H(t)=-9.8t^2+4.9t equation:

(H)=-9.8H^2+4.9H

We move all terms to the left:

(H)-(-9.8H^2+4.9H)=0

We get rid of parentheses

9.8H^2-4.9H+H=0

We add all the numbers together, and all the variables

9.8H^2-3.9H=0

a = 9.8; b = -3.9; c = 0;
Δ = b2-4ac
Δ = -3.92-4·9.8·0
Δ = 15.21
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{-(-3.9)-\sqrt{15.21}}{2*9.8}=\frac{3.9-\sqrt{15.21}}{19.6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3.9)+\sqrt{15.21}}{2*9.8}=\frac{3.9+\sqrt{15.21}}{19.6}
$