H(x)=3x^2+5x

Solution for H(x)=3x^2+5x equation:

(H)=3H^2+5H

We move all terms to the left:

(H)-(3H^2+5H)=0

We get rid of parentheses

-3H^2+H-5H=0

We add all the numbers together, and all the variables

-3H^2-4H=0

a = -3; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-3)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-3}=\frac{0}{-6}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-3}=\frac{8}{-6}
=-1+1/3
$