(x+2)(3x+2)=x^2+4x+4

Solution for (x+2)(3x+2)=x^2+4x+4 equation:

(x+2)(3x+2)=x^2+4x+4

We move all terms to the left:

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

We get rid of parentheses

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

We multiply parentheses ..

-x^2+(+3x^2+2x+6x+4)-4x-4=0

We add all the numbers together, and all the variables

-1x^2+(+3x^2+2x+6x+4)-4x-4=0

We get rid of parentheses

-1x^2+3x^2+2x+6x-4x+4-4=0

We add all the numbers together, and all the variables

2x^2+4x=0

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

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