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

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

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

We move all terms to the left:

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

We use the square of the difference formula

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

We multiply parentheses ..

x^2-((+2x^2+4x-4x-8))-4=0


We calculate terms in parentheses: -((+2x^2+4x-4x-8)), so:

(+2x^2+4x-4x-8)

We get rid of parentheses

2x^2+4x-4x-8

We add all the numbers together, and all the variables

2x^2-8

Back to the equation:

-(2x^2-8)


We get rid of parentheses

x^2-2x^2+8-4=0

We add all the numbers together, and all the variables

-1x^2+4=0

a = -1; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-1)·4
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{-(0)-4}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-1}=\frac{4}{-2}
=-2
$