OK either the equation is \(\sqrt{} 2x+4 - \sqrt{x} -2 =0\) or \(\sqrt{} 2x+4 = \sqrt{x} -2\) Either way it ends up the same because if it is the first equation, I would move the \(\sqrt{x} - 2\) to the other side of the equation. It would become \(\sqrt{x} + 2\).
For my example, I am going with \(\sqrt{} 2x+4 = \sqrt{x} -2\)
Square both sides
2x+4=\((x+2)^{2}\)
Then use FOIL to solve the right side
2x+4=\(x^{2} +2x+2x+4\)
Then combine like terms
2x+4=\(x^{2} +4x+4\)
Set it equal to 0 and combine like terms
\(0= x^{2} -2x\)
0=x(x-2)
So x=0 and x=2
If it was the other was it would be x=-2
Hope that helps.
