Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$. 
Solve the equation $13x + 2(3x + 2)\sqrt{x + 3} + 42 = 0$.

Let $y = \sqrt{x + 3} \implies 3 = y^2 - x$.
$$\large \begin{align}
&13x + 2(3x + 2)\sqrt{x + 3} +42\\
= &14(x + 3) + (6x + 4)y - x\\
= &14y^2 + [6(x + 3) - 14]y - x\\
= &14y(y - 1) - (y^2 - x - 9)y^3 - x\\
= &14y(y - 1) + x(y^3 - y) - y^3(y^2 - 1) + 8y^3\\
= &14y(y - 1) + (xy^2 + xy + x)(y - 1) - (y^4 + y^3)(y - 1) + 8y^3\\
= &(-y^4 + y^3 + xy^2 + xy + x + 14y)(y - 1) + 8y^3\\
\end{align}$$
And I'm stuck here.
 A: Part of the issue seems to be mixing up $x$ and $y$.  Usually once you identify a substitution it's easiest to bite the bullet and change the variable over completely.  Here, $y=\sqrt{x+3}$ implies $x=y^2-3$ and then
$13(y^2-3)+2(3y^2-7)y+42=0$
$6y^3+13y^2-14y+3=0$
Trying out rational root candidates with the Rational Root Theorem we identify $y=1/2$ as one of these roots.  Thereby
$(2y-1)(3y^2+8y-3)=0$
and solving the quadratic factor gives the additional roots $y=-3, y=1/3$.  The former is thrown out as the original substitution, by definition, requires a nonnegative square root.  Each of the other roots $y=1/2, y=1/3$ will give a valid solution for $x=y^2-3$ to the original equation.
A: I have no idea where you are trying to go with your approach, but I would suggest squaring both sides of
$$13x+42=-2(3x+2)\sqrt{x+3}.$$
This yields a cubic equation in $x$ which can be solved by standard methods; it turns out that all roots are rational so the rational root theorem yields them all. It remains to check whether these three roots are in fact solutions to the original equation.
A: Rearranging and squaring both sides of the original equation gives
$$4(3x+2)^2(x+3)=(13x+42)^2$$
$$36x^3+156x^2+160x+48=169x^2+1092x+1764$$
$$36x^3-13x^2-932x-1716=0$$
$$(x-6)(4x+11)(9x+26)=0$$
$$x=6, -\frac{11}{4}, -\frac{26}{9}$$
But $x=6$ is an extra solution created due to squaring both sides and will not work in the original equation.
