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?

To build off of the other answers provided, the equation you wish to solve is
$$\sqrt{2x-3} = 3-x$$
To do so, you square both sides, and solve
$$2x-3 = (3-x)^2$$
which has two solutions. However this equation can also be obtained by squaring 
$$-\sqrt{2x-3} = 3-x$$
The second solution is the solution to this second equation. This is easy to see with a plot:

