(x+1)2=(x+1)(3x+2)

Solution for (x+1)2=(x+1)(3x+2) equation:

(x+1)2=(x+1)(3x+2)

We move all terms to the left:

(x+1)2-((x+1)(3x+2))=0

We multiply parentheses

2x-((x+1)(3x+2))+2=0

We multiply parentheses ..

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


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

(+3x^2+2x+3x+2)

We get rid of parentheses

3x^2+2x+3x+2

We add all the numbers together, and all the variables

3x^2+5x+2

Back to the equation:

-(3x^2+5x+2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2-3x=0

a = -3; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-3}=\frac{0}{-6}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-3}=\frac{6}{-6}
=-1
$