How do we solve the equation? How do we solve the following equation in the set of real numbers? 
$$(x+1)\cdot \sqrt{x+2} + (x+6)\cdot \sqrt{x+7}=(x+3)\cdot (x+4).$$
I wrote the given equation has the form
\begin{equation*}
(x+1)(\sqrt{x  + 2} - 2) + (x + 6)(\sqrt{x+7} - 3) = (x-2)(x+4)
\end{equation*}
This  equation is equivalent to 
\begin{equation*}
(x-2)\left(\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4\right) = 0.
\end{equation*}
But I can not prove that the equation 
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 = 0
\end{equation*}
has no solution. Detail
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 <0, \forall x \geqslant -2. 
\end{equation*}

We have
$$\dfrac{x+1}{\sqrt{x+2}+2}<\dfrac{x+2}{\sqrt{x+2}+2}\leqslant \dfrac{x+2}{2}$$
and
$$\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+6}{3}\leqslant \dfrac{x+6}{2}.$$
Therefore
$$\dfrac{x+1}{\sqrt{x+2}+2}+\dfrac{x+6}{\sqrt{x+7}+3}<\dfrac{x+2}{2}+\dfrac{x+6}{2}=x+4$$
Thus the equation
\begin{equation*}
 \dfrac{x+1}{\sqrt{x+2} + 2} + \dfrac{x+6}{\sqrt{x+7} + 3} - x - 4 = 0
\end{equation*}
has no solution.
Thus, the given equation has only root $x = 2.$
