How do you solve this radical equation $\sqrt{2x+5} + 2\sqrt{x+6} = 5$? I have a radical equation 
$$
\sqrt{2x+5} + 2\sqrt{x+6} = 5
$$
and I am having trouble calculating an answer. I keep on getting weird numbers that are not correct as my answer. How do you solve this?  A step-by-step procedure would be highly appreciated.

First we require $x \geq -\frac{5}{2}$ (for both radicals to be well defined). Now we have that $\sqrt{2x+5}=5-2\sqrt{x+6}$. 
If $5-2\sqrt{x+6}\ge 0\Rightarrow -\frac{5}{2}\le x\le \frac{19}{4}$ then 
\begin{equation}\sqrt{2x+5}^2=(5-2\sqrt{x+6})^2\Leftrightarrow 2x+5=25+4(x+6)-20\sqrt{x+6}\end{equation}
Can you continue from here? Do you understand why we make the hypothesis?
Here is some of the rest:
\begin{equation} 2x+5=25+4(x+6)-20\sqrt{x+6}\Leftrightarrow 20\sqrt{x+6}=44+2x\Leftrightarrow 10\sqrt{x+6}=x+22\end{equation}
Since $x+22\ge 0$ (remember that $-\frac{5}{2}\le x\le \frac{19}{4}$) we have that
\begin{equation} (10\sqrt{x+6})^2=(x+22)^2\Leftrightarrow 100(x+6)=x^2+44x+484\Leftrightarrow x^2-56x-116=0\end{equation}
Solve the last equation and choose only the $x$ that lie in $\left[-\frac{5}{2}, \frac{19}{4}\right]$
