$\sqrt{2x+6}-\sqrt{x+4}=1$
solve for x.

Attempts:

$(\sqrt{2x+6}-\sqrt{x+4})\sqrt{2x+6}-\sqrt{x+4}=1(\sqrt{2x+6}-\sqrt{x+4})$
$2x+6+x+4=\sqrt{2x+6}-\sqrt{x+4}$

$2x+6+x+4+\sqrt{x+4}=\sqrt{2x+6}$
$2x+6=(3x+10)^2+x+4$
$2x+6=9x^2+61x+104$
$0=9x^2+59x+98$

$x=\frac{-59\pm\sqrt{47}i}{18}$

But, the answer sheet says that x=5. Any help?

2. $\begin{gathered}
\sqrt {2x + 6} = 1 + \sqrt {x + 4} \hfill \\
2x + 6 = 1 + 2\sqrt {x + 4} + x + 4 \hfill \\
x + 1 = 2\sqrt {x + 4} \hfill \\
x^2 + 2x + 1 = 4\left( {x + 4} \right) \hfill \\
\end{gathered}$

3. note that if the solution exists, then it must lie on $x\ge-3.$

write the inequality as $\sqrt{2x+6}=1+\sqrt{x+4},$ since both sides are positive, we can square them and get $2x+6=1+2\sqrt{x+4}+x+4$ then $x+1=2\sqrt{x+4},$ in order to square we require also $x\ge-1,$ and of course the previous condition, so we need $x\ge-1$ in order to make these things work, thus $x^2+2x+1=4x+16$ then $x^2-2x-15=0\implies(x+3)(x-5)=0,$ we require that $x\ge-1,$ so $x=5$ is the solution to our problem.