<s>[INST] Solving a mixed radical and quadratic equation 
Solve for $x \in \mathbb{R}$
$$4x^2(x+2) +3(2x^2-4x-3)\sqrt{4x+3} +6x = 0$$

I tried taking square by isolating the radical, but the resultant equation couldn't be solved.
Any help will be appreciated.
Thanks.
 [/INST] Let $X=4x+3$. Then,
$$3(2x^2-X)\sqrt{X}=-2x(2x^2+X)$$
Squaring the both sides gives
$$9(2x^2-X)^2X=4x^2(2x^2+X)^2,$$
i.e.
$$4x^4X-4x^2X^2+9X^3-16x^6+16x^4X-36x^2X^2=0$$
$$X(4x^4-4x^2X+9X^2)-4x^2(4x^4-4x^2X+9X^2)=0$$
$$(X-4x^2)(4x^4-4x^2X+9X^2)=0$$
since $9X^2=X^2+8X^2$
$$(X-4x^2)(4x^4-4x^2X+X^2+8X^2)=0$$
to have
$$(X-4x^2)((2x^2-X)^2+8X^2)=0$$
finally,
$$(-4x^2+4x+3)((2 x^2-4 x-3)^2+8 (4 x+3)^2)=0$$
and so
$$x=\frac 32,-\frac 12.$$
The former is sufficient while the latter isn't. So, $\color{red}{x=\frac 32}$ is the only solution.
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