H=-4.9t^2+8.8t+3

Solution for H=-4.9t^2+8.8t+3 equation:

=-4.9H^2+8.8H+3

We move all terms to the left:

-(-4.9H^2+8.8H+3)=0

We get rid of parentheses

4.9H^2-8.8H-3=0

a = 4.9; b = -8.8; c = -3;
Δ = b2-4ac
Δ = -8.82-4·4.9·(-3)
Δ = 136.24
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{-(-8.8)-\sqrt{136.24}}{2*4.9}=\frac{8.8-\sqrt{136.24}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8.8)+\sqrt{136.24}}{2*4.9}=\frac{8.8+\sqrt{136.24}}{9.8}
$