Solve the equation $\sqrt{3x-2} +2-x=0$ 
Solve the equation: $$\sqrt{3x-2} +2-x=0$$

I squared both equations $$(\sqrt{3x-2})^2  (+2-x)^2= 0$$ 
I got $$3x-2 + 4 -4x + x^2$$
I then combined like terms $x^2 -1x +2$
However, that can not be right since I get a negative radicand when I use the quadratic equation. 
$x = 1/2  \pm \sqrt{((-1)/2)^2 -2}$
The answer is 6

$$\sqrt{3x-2} +2-x=0$$

Isolating the radical:$$\sqrt{3x-2} =-2+x$$
Squaring both sides:$$\bigg(\sqrt{3x-2}\bigg)^2 =\bigg(-2+x\bigg)^2$$
Expanding $(-2+x)^2$ and gathering like terms: $$3x-2=-2(-2+x)+x(-2+x)$$
$$3x-2=4-2x-2x+x^2$$
Set x equal to zero:$$3x-2=4-4x+x^2$$
Gather like terms:$$0=4+2-3x-4x+x^2$$
Factor the quadratic and find the solutions:$$0=x^2-7x+6$$
$$0=(x-6)(x-1)$$
$$0=x-6\implies\boxed{6=x}$$ 
$$0=x-1\implies\boxed{1=x}$$

Checking 6 as a solution:
$$\sqrt{3(6)-2} +2-(6)=0$$
$$\sqrt{16} +2-6=0$$
$$4+2-6=0$$
$$6-6=0$$
Checking 1 as a solution:
$$\sqrt{3(1)-2} +2-(1)=0$$
$$\sqrt{1} +2-1=0$$
$$2\neq0$$
The solution $x=1$ does not equal zero and therefore is not a solution.
The solution $x=6$ does equal zero and therefore is our only solution to this equation.
