H(t)=-5t^2+5,5t

Solution for H(t)=-5t^2+5,5t equation:

(H)=-5H^2+5.5H

We move all terms to the left:

(H)-(-5H^2+5.5H)=0

We get rid of parentheses

5H^2-5.5H+H=0

We add all the numbers together, and all the variables

5H^2-4.5H=0

a = 5; b = -4.5; c = 0;
Δ = b2-4ac
Δ = -4.52-4·5·0
Δ = 20.25
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{-(-4.5)-\sqrt{20.25}}{2*5}=\frac{4.5-\sqrt{20.25}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4.5)+\sqrt{20.25}}{2*5}=\frac{4.5+\sqrt{20.25}}{10}
$