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.
 A: In fact my answer is the practical explanation of Old John's answer:
$$ [\sqrt{2x+5}+2 \sqrt{x+6}]^{2} = (2x+5)+4\sqrt{(2x+5)(x+6)} +4(x+6) = 25$$
$$ -4-6x = 4\sqrt{(2x+5)(x+6)}$$
$$ 16+48x+36{x}^{2} = 32{x}^{2}+272x+480$$
$$ x= -2 $$
$$ x= 58 $$
Then you check if they are solutions to the initial equation. You will get $x=-2$
A: 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 $\left[-\frac{5}{2}, \frac{19}{4}\right]$
A: Step 1: square both sides, possibly after moving one of the square roots to the other side.
You should now have an equation with only one square root in it.
Step 2: isolate the remaining square root on its own on one side.
Step 3: square both sides to get rid of the remaining square root.
Step 4: solve the polynomial equation you now have.
Step 5: check which of the solutions you obtained in step 4 really are solutions of the original equation. This step is necessary , since squaring both sides at various steps will have introduced extra solutions which are not solutions of the equation you started with.
A: $\rm\begin{eqnarray}\rm  a^2 &=&\rm 2x\!+\!5 \\ \rm b^2 &=&\rm\ \: x\!+\!6 \end{eqnarray}\:\Rightarrow\: a^2\!-\!2b^2 = -7.\ $ But  $\rm\:\color{#C00}a\!+\!2b = 5\:$ so $\rm\:0 = (\color{#C00}{5\!-\!2b})^2\!-2b^2\!+\!7\, =\, 2\,(b\!-\!8)\,(b\!-\!2)\:$
A: Left hand side of the equation is increasing, so there is at most one root. 
Careful gazing gives $x=-2$.
