Is the problem which method to use or using the methods? In general, if you can't figure out how to solve a quadratic, there's always the quadratic formula to fall back on. (This is going to look terrible without LaTeX, sorry!)

ax^2 + bx + c = 0 implies

x = -b/(2a) (+/-) [sqrt(b^2 - 4ac)]/(2a) (Where the "+/-" is "plus or minus.")

The method of "completing the square" always works as well. Then there's the methods you can use to factor, but at a glance I don't think any of these factor.

I'll do this one by completing the square, if you need help with the quadratic formula, just let me know.

x^2 - 6x + 3 = 0

x^2 - 6x = -3

We want to add a number to both sides such that the LHS is a perfect square. The format for this is the identity:

(x + a)^2 = x^2 + (2a)x + a^2.

So, looking at the form (with the "a" in it) and your problem the coefficient of the x term is - 6 = 2a. Thus a = -3. So we want to add (-3)^2 to both sides of the equation:

x^2 - 6x + 9 = -3 + 9

(x - 3)^2 = 6

x - 3 = (+/-)sqrt(6)

x = 3 (+/-) sqrt(6).

-Dan