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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(3(x+1)(x+1)), so:

3(x+1)(x+1)

We multiply parentheses ..

3(+x^2+x+x+1)

We multiply parentheses

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

We add all the numbers together, and all the variables

3x^2+6x+3

Back to the equation:

-(3x^2+6x+3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-2x-2=0

a = 1; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·1·(-2)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2*1}=\frac{2-2\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2*1}=\frac{2+2\sqrt{3}}{2}
$