3x^{2 }-12x + 6 = 0

I divide by 3 to get

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

then I subtract 2 from each side

x^{2 }-4x = -2

(x-2)^{2} = -2

x-2 = -2^{1/2
}x= 2+- (-2^{1/2)}

but the book shows the answer to be +2^{1/2} NOT -2^{1/2} ...where did I go wrong?

As already pointed out, x

^{2 }-4x \(\displaystyle \neq\)(x-2)

^{2}. You need to fix that. I however want to talk about what you said after that.

You wrote x-2 = -2

^{1/2}. Now I agree that the right side is -2 and agree that you should take the square root of it. The problem is that the sqrt of -2 is NOT -(sqrt(2)). You did not take the square root of the entire right side but you did take the square root of the entire left side. The sqrt of the right side is (-2)

^{1/2 }or sqrt(-2), Not -(sqrt(2).

You do now that -(sqrt(4)) and sqrt(-4) are different? Since sqrt(4) = 2, we have -(sqrt(4)) = -(2) = -2 while sqrt(-4) has no real solution.

The next thing I want to point out is where you went (again) from (x-2)

^{2} = -2 to x-2 = -2

^{1/2}. If (x-2)

^{2} = -2, then you should say that there is no answer since Nothing squared is ever a negative number (like -2!)

Also I doubt that your book says that x = 2

^{1/2} as that is not any of the solution. Please reply back showing your corrected work so that we can help you or congratulate you for getting the correct result.