H(t)=-5t2+10t

Solution for H(t)=-5t2+10t equation:

(H)=-5H^2+10H

We move all terms to the left:

(H)-(-5H^2+10H)=0

We get rid of parentheses

5H^2-10H+H=0

We add all the numbers together, and all the variables

5H^2-9H=0

a = 5; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·5·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*5}=\frac{0}{10}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*5}=\frac{18}{10}
=1+4/5
$