H(t)=-4.9t^2+8t=0

Solution for H(t)=-4.9t^2+8t=0 equation:

(H)=-4.9H^2+8H=0

We move all terms to the left:

(H)-(-4.9H^2+8H)=0

We get rid of parentheses

4.9H^2-8H+H=0

We add all the numbers together, and all the variables

4.9H^2-7H=0

a = 4.9; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·4.9·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*4.9}=\frac{0}{9.8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*4.9}=\frac{14}{9.8}
=1+3.5/8.1666666666667
$