H(t)=-4.9t^2+14t+2=0

Solution for H(t)=-4.9t^2+14t+2=0 equation:

(H)=-4.9H^2+14H+2=0

We move all terms to the left:

(H)-(-4.9H^2+14H+2)=0

We get rid of parentheses

4.9H^2-14H+H-2=0

We add all the numbers together, and all the variables

4.9H^2-13H-2=0

a = 4.9; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·4.9·(-2)
Δ = 208.2
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{-(-13)-\sqrt{208.2}}{2*4.9}=\frac{13-\sqrt{208.2}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{208.2}}{2*4.9}=\frac{13+\sqrt{208.2}}{9.8}
$