<s>[INST] Solve the equation $4x^3+13x^2-14x=3-\sqrt{15x+9}$ I ran into this problem:
$$4x^3+13x^2-14x=3-\sqrt{15x+9}$$
It makes no sense to square it. I thought it was necessary to make a replacement. What kind of substitution?
First of all, the root gets in the way, so you have to make a substitution that removes the root!
I was able to find such a substitution
$$x=\frac{9}{15}\cos 2t\Rightarrow \sqrt{15x+9}=3\sqrt{2}\cos t$$
We don't have to put the module, because the restrictions allow us to do so
After we got rid of the root, then comes the second problem. How do we solve this equation?
$$\frac{108}{125}\cos^32 t+\frac{117}{25}\cos^22t-\frac{42}{5}\cos 2t=3-3\sqrt{2}\cos t$$
 [/INST] The LHS of the equation
$$4x^3+13x^2-14x=3-\sqrt{15x+9}$$
equals $0$ when $x=0$ for which the RHS also equals $0$.
So, here, I would add $ax$ to the both sides (each side still equals $0$ when $x=0$) since after having
$$4x^3+13x^2-14x+ax=ax+3-\sqrt{15x+9}$$
$$4x^3+13x^2-14x+ax=\frac{(ax+3)^2-(15x+9)}{ax+3+\sqrt{15x+9}}$$
$$x(4x^2+13x-14+a)=\frac{x(a^2x+6a-15)}{ax+3+\sqrt{15x+9}}\tag1$$
it may be possible to have something like
$$x(x-c)(\text{something positive})=0$$
if the both sides of $(1)$ have a common factor $(x-c)$.
This means that the solution of $a^2x+6a-15=0$ has to be a solution of $4x^2+13x-14+a=0$, which means that $a$ has to satisfy
$$4\bigg(\frac{-6a+15}{a^2}\bigg)^2+13\bigg(\frac{-6a+15}{a^2}\bigg)-14+a=0$$
i.e.
$$\frac{a^5 - 14 a^4 - 78 a^3 + 339 a^2 - 720 a + 900}{a^4} = 0$$
Fortunately the rational root theorem works to get $a=2$.

Solution :
$$\begin{align}&4x^3+13x^2-14x=3-\sqrt{15x+9}
\\\\&\iff 4x^3+13x^2-14x+2x=2x+3-\sqrt{15x+9}
\\\\&\iff x(4x-3)(x+4)=\frac{(2x+3)^2-(15x+9)}{2x+3+\sqrt{15x+9}}
\\\\&\iff x(4x-3)(x+4)=\frac{x(4x-3)}{2x+3+\sqrt{15x+9}}
\\\\&\iff x(4x-3)\bigg(x+4-\frac{1}{2x+3+\sqrt{15x+9}}\bigg)=0
\\\\&\iff x(4x-3)\bigg(x+1+3-\frac{1}{2x+3+\sqrt{15x+9}}\bigg)=0
\\\\&\iff x(4x-3)\bigg(\underbrace{x+1+\frac{6(x+1)+2+3\sqrt{15x+9}}{2(x+1)+1+\sqrt{15x+9}}}_{\text{positive since $x\geqslant -9/15\gt -1$}}\bigg)=0
\\\\&\iff x=0,\frac 34\end{align}$$
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