How can I solve this nice rational equation I am trying solve this equation
$$\dfrac{3x^2 + 4x + 5}{\sqrt{5x^2 + 4x +3}}+\dfrac{8x^2 + 9x + 10}{\sqrt{10x^2 + 9x +8}} = 5.$$ Where $x \in \mathbb{R}$. I knew that $x=-1$ is a given solution. But I can't solve it. I tried
We rewrite the given equation in the form
$$\dfrac{3(x^2 + x + 1) + x + 2}{\sqrt{5(x^2 + x + 1) - (x+2)}}+\dfrac{8(x^2 + x + 1) + x + 2}{\sqrt{10(x^2 + x + 1)- (x+2)} } = 5.$$
Put $a = x^2 + x + 1$ and $b = x + 2$, we get
$$\dfrac{3a  + b}{\sqrt{5a - b}}+\dfrac{8a + b}{\sqrt{10a- b} } = 5.$$
From here, I stoped.

The equation of tangent line of the grap of the funtion  $ y = \dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}$ at the point $x=-1$ is $y=\dfrac{x}{2}+\dfrac{5}{2}$ and the equation of tangent line of the grap of the funtion $ y = \dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}$ at the point $x=-1$ is $y=\dfrac{5}{2}-\dfrac{x}{2}$. We prove that
\begin{equation}
\label{eq4_30_06_2016}
\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}} \geqslant \dfrac{x}{2}+\dfrac{5}{2}
\end{equation}
and
\begin{equation}
\label{eq5_30_06_2016}
\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}} \geqslant \dfrac{5}{2}-\dfrac{x}{2}.
\end{equation}
The first inequality  equavalent to 
\begin{equation}
\label{eq1_30_06_2016}
2(3 x^2+4 x+5) \geqslant (x+5)\sqrt{5 x^2+4 x+3}.
\end{equation}
If $x+5 \leqslant 0$, it is always true.
If $x+5 > 0$, squaring  both sides , we get
$$31 x^4+42 x^3+16 x^2+30 x+25 \geqslant 0.$$
Equavalent to
\begin{equation}
\label{eq_2_30_06_2016}
(x+1)^2\cdot \left(31 x^2-20 x+25\right) \geqslant 0.
\end{equation}
Similarly, if $x \geqslant 5$ the second equality is true. If $x < 5$, squaring both sides, we get
$$(x+1)^2 \cdot\left(246 x^2+175 x+200\right) \geqslant 0.$$
Equality of two equalities hold iff $x = -1$.
Add two equalities, we have
$$\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}+\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}\geqslant 5.$$
Therefore, the equation 
$$\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}+\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}= 5$$
has only solution $x=-1.$
