$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$.

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$2.59 \sqrt{x+2}-\sqrt[3]{3 x+2}=0$.
2.59 We have $\sqrt{x+2}=\sqrt[3]{3 x+2}$. After raising the equation to the sixth power, we get $(x+2)^{3}=(3 x+2)^{2}$, or $x^{3}-3 x^{2}+4=0$, or $(x+1)\left(x^{2}-4 x+4\right)=0$.

From this, $x_{1}=-1$ (extraneous root), $x_{2}=2$.

Answer: $x=2$.