Please help me solve this $x(\sqrt{2x+5}+\sqrt[3]{7x+13}) = 3x+6$

Question

Wolfram Alpha shows that the answer is $x=2\,$ and $x=-2\,$ but what would be the best way of simplifying this equation ? It has been many years since I was in school , and I just cannot wrap my head around it. :/

Answer 1

$$x(\sqrt{2x+5}+\sqrt[3]{7x+13})=3x+6$$ $x\not=0$ $$\sqrt{2x+5}+\sqrt[3]{7x+13}=\frac{3x+6}x$$ $$\sqrt{2x+5}+\sqrt[3]{7x+13}=3+\frac6x$$

$x \in [-\frac52;0)\cup(0;+\infty)$

1) $x\in [-\frac52;0)$

Then $f(x)=\sqrt{2x+5}+\sqrt[3]{7x+13} -$ increasing function; $g(x)=3+\frac6x -$ decreasing function. Then $x=-2 -$ only root

2) $x\in (0;+\infty)$

Then $f(x)=\sqrt{2x+5}+\sqrt[3]{7x+13} -$ increasing function; $g(x)=3+\frac6x -$ decreasing function. Then $x=2 -$ only root

Addition:

Find the roots by hand and shows these two are the only roots. For example:

$$\sqrt[3]{x-2}+\sqrt{x-1}=5$$