H(t)=13.7t+.5-4.9t^2

Solution for H(t)=13.7t+.5-4.9t^2 equation:

(H)=13.7H+.5-4.9H^2

We move all terms to the left:

(H)-(13.7H+.5-4.9H^2)=0

We get rid of parentheses

4.9H^2-13.7H+H-.5=0

We add all the numbers together, and all the variables

4.9H^2-12.7H-0.5=0

a = 4.9; b = -12.7; c = -0.5;
Δ = b2-4ac
Δ = -12.72-4·4.9·(-0.5)
Δ = 171.09
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{-(-12.7)-\sqrt{171.09}}{2*4.9}=\frac{12.7-\sqrt{171.09}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12.7)+\sqrt{171.09}}{2*4.9}=\frac{12.7+\sqrt{171.09}}{9.8}
$