H=-0.5t^2+3t+2

Solution for H=-0.5t^2+3t+2 equation:

=-0.5H^2+3H+2

We move all terms to the left:

-(-0.5H^2+3H+2)=0

We get rid of parentheses

0.5H^2-3H-2=0

a = 0.5; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·0.5·(-2)
Δ = 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{-(-3)-\sqrt{13}}{2*0.5}=\frac{3-\sqrt{13}}{1}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{13}}{2*0.5}=\frac{3+\sqrt{13}}{1}
$