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*}

Not much simpler, but you can also try something like $a:=\sqrt{x+2}$ then $\sqrt{x+7}=\sqrt{a^2+5}$, then it will have only one sqrt in the equation, put that on one side and the rest on the other side.. 
$$(a^2-1)a+(a^2+4)\cdot\sqrt{a^2+5} = (a^2+1)(a^2+3)$$
