Finding all real roots of the equation $(x+1) \sqrt{x+2} + (x+6)\sqrt{x+7} = x^2+7x+12$

We have $$x^2+7x+12-(x+1) \sqrt{x+2} - (x+6)\sqrt{x+7}=0$$ Multiplying the both sides by $3$ gives $$3x^2+21x+36-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0\tag1$$ Now since $$\begin{align}&3x^2+21x+36\\&=x^2+5x+4+x^2+13x+42+x^2+3x-10\\&=(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)\end{align}$$ we have, from $(1)$, $$(x+1)(x+4)+(x+6)(x+7)+(x-2)(x+5)-3(x+1) \sqrt{x+2} - 3(x+6)\sqrt{x+7}=0$$ Rearranging terms $$(x+1)(x+4-3\sqrt{x+2})+(x+6)\sqrt{x+7}\ (\sqrt{x+7}-3)+(x-2)(x+5)=0,$$ i.e. $$(x+1)\cdot\frac{(x-2)(x+1)}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}\ (x-2)}{\sqrt{x+7}+3}+(x-2)(x+5)=0$$ and so $$(x-2)f(x)=0$$ where $$f(x)=\frac{(x+1)^2}{x+4+3\sqrt{x+2}}+\frac{(x+6)\sqrt{x+7}}{\sqrt{x+7}+3}+(x+5)$$ We know that $f(x)$ is positive because of $x\ge -2$.

Thus, $\color{red}{x=2}$ is the only solution.