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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

0.5H^2-2H-2=0

a = 0.5; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·0.5·(-2)
Δ = 8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*0.5}=\frac{2-2\sqrt{2}}{1}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*0.5}=\frac{2+2\sqrt{2}}{1}
$