Question 297948
We have (x-6)*(x+6) = x*x - 36 .  I would say x is all real numbers because this equation is simply an identity equation.  (x-6)*(x+6) is equal to x*x - 36.  We are dealing with a difference of squares.

let me expand it more slowly:  (x-6)*(x+6) = x*x + 6x - 6x + (-6)*6 = x*x - 36

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However if your notation x2 really meant x times 2 instead of x squared, then x has a finite set of solutions.

Let's assume (x-6)(x+6) = 2*x -36,  then we have

x*x - 36 = 2x - 36
add 36 to both sides to cancel and we have:
x*x = 2x
x = 0 or x = 2

Sincerely Alan C.
CSU Math Graduate 07