Why do I keep getting this incorrect solution when trying to find all the real solutions for $\sqrt{2x-3}\ +x=3$. The problem is to find all real solutions (if any exists) for $\sqrt{2x-3}\ +x=3$.
Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:
I got rid of the root by squaring both sides, 
$$\sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got 
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, \ x=6$$
Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?

Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.
As a very simple example, notice the following two equations:
$$x = \sqrt 4 \iff x = +2$$
$$x^2 = 4 \iff \vert x\vert = 2 \iff x = \pm 2$$
The first equation has only one solution: $+\sqrt 4$. The second, however, has two solutions: $\pm\sqrt 4$. And you get the second equation by squaring the first one.
The exact same idea applies to your example. You have
$$\sqrt{2x-3} = 3-x$$
which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving
$$0 = 12-8x+x^2 \iff \color{blue}{\pm}\sqrt{2x-3} = 3-x$$
which has a $\pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:
$$2x-3 = 9-6x+x^2; \quad \color{blue}{x \leq 3}$$
$$0 = 12-8x+x^2; \quad \color{blue}{x \leq 3}$$
Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.
