H(t)=-5t2+10t+1,5

Solution for H(t)=-5t2+10t+1,5 equation:

(H)=-5H^2+10H+1.5

We move all terms to the left:

(H)-(-5H^2+10H+1.5)=0

We get rid of parentheses

5H^2-10H+H-1.5=0

We add all the numbers together, and all the variables

5H^2-9H-1.5=0

a = 5; b = -9; c = -1.5;
Δ = b2-4ac
Δ = -92-4·5·(-1.5)
Δ = 111
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{-(-9)-\sqrt{111}}{2*5}=\frac{9-\sqrt{111}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{111}}{2*5}=\frac{9+\sqrt{111}}{10}
$