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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4.9H^2+H-7=0

a = 4.9; b = 1; c = -7;
Δ = b2-4ac
Δ = 12-4·4.9·(-7)
Δ = 138.2
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{-(1)-\sqrt{138.2}}{2*4.9}=\frac{-1-\sqrt{138.2}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{138.2}}{2*4.9}=\frac{-1+\sqrt{138.2}}{9.8}
$