H(n)=3n2-4

Solution for H(n)=3n2-4 equation:

(H)=3H^2-4

We move all terms to the left:

(H)-(3H^2-4)=0

We get rid of parentheses

-3H^2+H+4=0

a = -3; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·(-3)·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-3}=\frac{-8}{-6}
=1+1/3
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-3}=\frac{6}{-6}
=-1
$