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

Solution for 2(x+1)(x-1)=(2x+4)(x+3) equation:

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

We move all terms to the left:

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

We use the square of the difference formula

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

We multiply parentheses ..

x^2-((+2x^2+6x+4x+12))-1=0


We calculate terms in parentheses: -((+2x^2+6x+4x+12)), so:

(+2x^2+6x+4x+12)

We get rid of parentheses

2x^2+6x+4x+12

We add all the numbers together, and all the variables

2x^2+10x+12

Back to the equation:

-(2x^2+10x+12)


We get rid of parentheses

x^2-2x^2-10x-12-1=0

We add all the numbers together, and all the variables

-1x^2-10x-13=0

a = -1; b = -10; c = -13;
Δ = b2-4ac
Δ = -102-4·(-1)·(-13)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{3}}{2*-1}=\frac{10-4\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{3}}{2*-1}=\frac{10+4\sqrt{3}}{-2}
$