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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(+2x^2-4x)

We get rid of parentheses

2x^2-4x

Back to the equation:

-(2x^2-4x)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+4x=0

a = -1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-1)·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*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-1}=\frac{0}{-2}
=0
$