H=-4.9t^2+40t+6

Solution for H=-4.9t^2+40t+6 equation:

=-4.9H^2+40H+6

We move all terms to the left:

-(-4.9H^2+40H+6)=0

We get rid of parentheses

4.9H^2-40H-6=0

a = 4.9; b = -40; c = -6;
Δ = b2-4ac
Δ = -402-4·4.9·(-6)
Δ = 1717.6
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{-(-40)-\sqrt{1717.6}}{2*4.9}=\frac{40-\sqrt{1717.6}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+\sqrt{1717.6}}{2*4.9}=\frac{40+\sqrt{1717.6}}{9.8}
$