H(t)=10+2t-4.9t^2

Solution for H(t)=10+2t-4.9t^2 equation:

(H)=10+2H-4.9H^2

We move all terms to the left:

(H)-(10+2H-4.9H^2)=0

We get rid of parentheses

4.9H^2-2H+H-10=0

We add all the numbers together, and all the variables

4.9H^2-1H-10=0

a = 4.9; b = -1; c = -10;
Δ = b2-4ac
Δ = -12-4·4.9·(-10)
Δ = 197
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{-(-1)-\sqrt{197}}{2*4.9}=\frac{1-\sqrt{197}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{197}}{2*4.9}=\frac{1+\sqrt{197}}{9.8}
$