H(t)=-4.9t^2+29.4t+8

Solution for H(t)=-4.9t^2+29.4t+8 equation:

(H)=-4.9H^2+29.4H+8

We move all terms to the left:

(H)-(-4.9H^2+29.4H+8)=0

We get rid of parentheses

4.9H^2-29.4H+H-8=0

We add all the numbers together, and all the variables

4.9H^2-28.4H-8=0

a = 4.9; b = -28.4; c = -8;
Δ = b2-4ac
Δ = -28.42-4·4.9·(-8)
Δ = 963.36
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{-(-28.4)-\sqrt{963.36}}{2*4.9}=\frac{28.4-\sqrt{963.36}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28.4)+\sqrt{963.36}}{2*4.9}=\frac{28.4+\sqrt{963.36}}{9.8}
$