Solve $\sqrt{7-3x}+3=x$ I have the equation $\sqrt{7-3x}+3=x$ and I need to find $x$.
I eventually found the answer to be $1$ and $2$. 
In order to check my answer, I plugged $1$ first back into the equation.
I eventually got to the form: $\sqrt 4 +3 =1$. I know that the square root of $4$ has two answers: $-2$ and $2$. If used $-2$, the would count as a solution. If I use $2$, the solution would be considered extraneous. I was confused add search it up on wolframalpha, which stated that there are no answers. I was if the solution was extraneous or not and the reasoning behind it.  
 A: There is no solution to this equation. When you square it and solve the quadratic you get $x=1$ or $x=2$ as you have seen. But this does not guarantee that these are necessarily solutions. When you go back to the equation you finds that these do not satisfy the equation. So there is no solution. 
A: In the reals, the square root of a number is defined to be positive, hence $\sqrt4=-2$ does not hold.

A rigorous resolution is
$$\begin{align}\sqrt{7-3x}+3=x&\iff \sqrt{7-3x}=x-3\\&\iff 7-3x=(x-3)^2\color{green}{\land x-3\ge0}
\\&\iff x^2-3x+2=0\land x\ge3
\\&\iff (x=1\lor x=2)\land x\ge3.\end{align}$$
A: Here is a rough argument to verify there are no solutions:
The domain of $\sqrt{7-3x} + 3$ is $[-\infty, \frac{7}{3}]$. Since the function is continuous, from the shape of the square root function, the minimum value is at $x=  \frac{7}{3}$. This can also be proved more rigorously by differentiating and showing that the derivative is always less than $0$, so the minimum must be attained at the endpoint.
However, since in the domain of $\sqrt{7-3x} + 3$, $x$ never exceeds $\sqrt{7-3 \cdot \frac{7}{3}} + 3 = 3$, there are no solutions.
A: Set $y=\sqrt{7-3x}\ge0$
$\implies y^2=7-3x\implies3x=7-y^2$
$$\sqrt{7-3x}+3=x\implies3(y+3)=7-y^2$$
$$\implies y^2+3y+2=0, y=-1\text{ or -3}$$
But $y\ge0$
Alternatively,  $y\ge0\implies y^2+3y+2\ge0+0+2\ge2>0$
