<s>[INST] 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
 [/INST] What you did was not squaring both sides of the equation. If we have an equation
$$a+b=0,$$
then squaring both sides produces
$$\begin{align*}
(a+b)^2&=0^2\\a^2+2ab+b^2&=0.
\end{align*}$$
In your case, you have $a=\sqrt{3x-2}$ and $b=2-x$. But when you squared the left side of the equation, what you've written is instead $a^2+b^2$. But 
$$(a+b)^2=a^2+2ab+b^2\neq a^2+b^2.$$
Thus, when correctly squaring both sides of the equation you've been given, the result is
$$(\sqrt{3x-2})^2 +2(\sqrt{3x-2})(2-x)+(2-x)^2=0.$$
However, this is not a productive place to start solving the question. A better approach would be to move the things not inside the radical to the other side, and then square, like this:
$$\begin{align*}
\sqrt{3x-2} +2-x&=0\\\\
\sqrt{3x-2} &=x-2\\\\
\qquad\qquad\qquad(\sqrt{3x-2})^2&=(x-2)^2\qquad \leftarrow \text{squared both sides}
\end{align*}$$
Can you take it from here?
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