H=-4.9t^2+20t+4

Solution for H=-4.9t^2+20t+4 equation:

=-4.9H^2+20H+4

We move all terms to the left:

-(-4.9H^2+20H+4)=0

We get rid of parentheses

4.9H^2-20H-4=0

a = 4.9; b = -20; c = -4;
Δ = b2-4ac
Δ = -202-4·4.9·(-4)
Δ = 478.4
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{-(-20)-\sqrt{478.4}}{2*4.9}=\frac{20-\sqrt{478.4}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{478.4}}{2*4.9}=\frac{20+\sqrt{478.4}}{9.8}
$