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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

0.5H^2-1H-3=0

a = 0.5; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·0.5·(-3)
Δ = 7
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{-(-1)-\sqrt{7}}{2*0.5}=\frac{1-\sqrt{7}}{1}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{7}}{2*0.5}=\frac{1+\sqrt{7}}{1}
$