H(t)=-3X2+2,5

Solution for H(t)=-3X2+2,5 equation:

(H)=-3H^2+2.5

We move all terms to the left:

(H)-(-3H^2+2.5)=0

We get rid of parentheses

3H^2+H-2.5=0

a = 3; b = 1; c = -2.5;
Δ = b2-4ac
Δ = 12-4·3·(-2.5)
Δ = 31
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{-(1)-\sqrt{31}}{2*3}=\frac{-1-\sqrt{31}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{31}}{2*3}=\frac{-1+\sqrt{31}}{6}
$