H=-0.5t^2+3t

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

=-0.5H^2+3H

We move all terms to the left:

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

We get rid of parentheses

0.5H^2-3H=0

a = 0.5; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·0.5·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*0.5}=\frac{0}{1}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*0.5}=\frac{6}{1}
=6
$