<s>[INST] Solve the equation: $(9x^2+6x-8)\sqrt{3x+2}+6x+23=27x^2+3\sqrt{10+3x}$ Solve the equation:
$(9x^2+6x-8)\sqrt{3x+2}+6x+23=27x^2+3\sqrt{10+3x}$

I used wolframalpha.com and got only solution $x=-\dfrac{1}{3}$.
And this is my try:

Condition: $x\ge-\dfrac{2}{3}$.
The equation is equivalent to:
$\quad(3x+1)^2\sqrt{3x+2}+9-9\sqrt{3x+2}=27x^2-6x-5+3\sqrt{10+3x}-9$
$\Leftrightarrow (3x+1)^2\sqrt{3x+2}-\dfrac{9(3x+1)}{1+\sqrt{3x+2}}=(3x+1)(9x-5)+\dfrac{3(3x+1)}{\sqrt{10+3x}+3}$
$\Leftrightarrow (3x+1)\left[(3x+1)\sqrt{3x+2}-\dfrac{9}{1+\sqrt{3x+2}}-9x+5-\dfrac{3}{\sqrt{10+3x}+3}\right]=0$

But I can't prove that $(3x+1)\sqrt{3x+2}-\dfrac{9}{1+\sqrt{3x+2}}-9x+5-\dfrac{3}{\sqrt{10+3x}+3}>0$.
So, who can help me solve it?
 [/INST] I think the polynomial that you get by multiplying by the three conjugates is 
$$59049x^{10}-118098x^9+164025x^8-236196x^7 -441774x^6+133164x^5+373086x^4-390636x^3 -193203x^2+141054x+50641=0$$
This has four real zeros, including $x=-1/3$, but those would include negative signs for the square-roots.
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