H=-2t2+3t+2

Solution for H=-2t2+3t+2 equation:

=-2H^2+3H+2

We move all terms to the left:

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

We get rid of parentheses

2H^2-3H-2=0

a = 2; b = -3; c = -2;
Δ = b2-4ac
Δ = -32-4·2·(-2)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*2}=\frac{-2}{4}
=-1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*2}=\frac{8}{4}
=2
$