Question 520627
First find where it's equal to 0. These points will determine the bounds of the set of numbers we want to satisfy the question.

x^2-6x+9=0 can be factored and is thus equivalent to (x-3)^2=0

So when x = 3, the function goes to 0. Let's pick a number greater than 3, and one lower than 3 and see what happens to the given function. 

For x = 4; (x-3)^2 = (4-3)^2 = 1, which is greater than 0. In fact if x>3, then (x-3)>1, which means for all x>3, (x-3)^2 > 0. So no x>3 will be part of our solution.

For x = 2; (x-3) < 0, however, (x-3)^2 > 0, since any number squared becomes positive. So it turns out for x<3, (x-3)^2 > 0 as well.

So ultimately, the only solution is x=3, since at x = 3, x^2-6x+9=0. For all other x, x^2-6x+9>0.