<s>[INST] Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x+2}=0\\(x+1)(x+2)-2x\sqrt{3x+2}=0$$ What next? Thank you!
 [/INST] Multiply by $\sqrt{3x+2}$, assuming $3x+2\ge 0$ i.e. $x\ge-2/3$:
$$\left(\sqrt{3x+2}\right)^2-2x\sqrt{3x+2}+x^2=0$$
$$\left(\sqrt{3x+2}-x\right)^2=0$$
$$\sqrt{3x+2}=x$$
$$3x+2=x^2$$
Solve the quadratic equation to find $x=\frac{3\pm\sqrt{17}}{2}$. As the final equation is a consequence of the original one, but not necessarily equivalent to it, you now need to try those two solutions in the original equation. The only real solution will turn out to be $x=\frac{3+\sqrt{17}}{2}$ : the other one will be spurious and needs to be rejected.
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