H(t)=-4.9t^2-4t+3

Solution for H(t)=-4.9t^2-4t+3 equation:

(H)=-4.9H^2-4H+3

We move all terms to the left:

(H)-(-4.9H^2-4H+3)=0

We get rid of parentheses

4.9H^2+4H+H-3=0

We add all the numbers together, and all the variables

4.9H^2+5H-3=0

a = 4.9; b = 5; c = -3;
Δ = b2-4ac
Δ = 52-4·4.9·(-3)
Δ = 83.8
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{-(5)-\sqrt{83.8}}{2*4.9}=\frac{-5-\sqrt{83.8}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{83.8}}{2*4.9}=\frac{-5+\sqrt{83.8}}{9.8}
$