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

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

(H)=(2H)(2H+3)

We move all terms to the left:

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


We calculate terms in parentheses: -(2H(2H+3)), so:

2H(2H+3)

We multiply parentheses

4H^2+6H

Back to the equation:

-(4H^2+6H)


We get rid of parentheses

-4H^2+H-6H=0

We add all the numbers together, and all the variables

-4H^2-5H=0

a = -4; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-4)·0
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{-(-5)-5}{2*-4}=\frac{0}{-8}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-4}=\frac{10}{-8}
=-1+1/4
$