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 = 5
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{5}}{2*-0.5}=\frac{3-\sqrt{5}}{-1}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{5}}{2*-0.5}=\frac{3+\sqrt{5}}{-1}
$