H=2t^2-3t+0.5

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

=2H^2-3H+0.5

We move all terms to the left:

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

We get rid of parentheses

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

a = -2; b = 3; c = -0.5;
Δ = b2-4ac
Δ = 32-4·(-2)·(-0.5)
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*-2}=\frac{-3-\sqrt{5}}{-4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{5}}{2*-2}=\frac{-3+\sqrt{5}}{-4}
$