H=O.3+5.5t-4.9t^2

Solution for H=O.3+5.5t-4.9t^2 equation:

=.3+5.5H-4.9H^2

We move all terms to the left:

-(.3+5.5H-4.9H^2)=0

We get rid of parentheses

4.9H^2-5.5H-.3=0

We add all the numbers together, and all the variables

4.9H^2-5.5H-0.3=0

a = 4.9; b = -5.5; c = -0.3;
Δ = b2-4ac
Δ = -5.52-4·4.9·(-0.3)
Δ = 36.13
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{-(-5.5)-\sqrt{36.13}}{2*4.9}=\frac{5.5-\sqrt{36.13}}{9.8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5.5)+\sqrt{36.13}}{2*4.9}=\frac{5.5+\sqrt{36.13}}{9.8}
$