Can someone help with an algebra problem?

[HSmom25andRN]

Jacob's algebra chapter 13 lesson 6 problem 9f. The problem is

X squared - 6X - 9 = 0

The answer in the back of the book is 3 + 3 times the square root of 2 and 3- 3 times the square root of 2.

I figured out all of the other problems but this one has me stumped.

[creekland]

Using the quadratic equation: x= (-b +/- square root b squared - 4ac)/ 2a

a = 1
b = -6
c = -9

So,

x = (-(-6) +/- square root 36 - (4*1*-9) / 2*1

x = (6 +/- square root 36 + 36)/ 2

x = (6 +/- square root of 72) / 2

x = (6 +/ square root 36*2) /2 editing to add... break the 72 up into 36 * 2 in order to pull out the 6 (we can easily take the square root of 36, but not of 2 or 72. Taking out what we can simplifies it.)

x = (6 +/ 6(square root 2))/ 2

divide both terms by 2 and

x = 3 + 3 square root 2 or x = 3 - 3 square root 2

[Dee in MI]

I used the quadratic formula and matched the book. But here is how Wolfram|Alpha solved it:

http://www.wolframalpha.com/input/?i=solve+x^2-6x-9

Hit the orange "Show steps" below the results to get their step-by-step solution.

[HSmom25andRN]

My answer was X = (+/- the square root of 18) plus 3

so I just forgot about simplifying the square root of 18

I was going crazy. You guys are the best. Thanks you.