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
$