H(x)=3x2+2x-4

Solution for H(x)=3x2+2x-4 equation:

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

We move all terms to the left:

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

We get rid of parentheses

-3H^2+H-2H+4=0

We add all the numbers together, and all the variables

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