<s>[INST] 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.
 [/INST] 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.
</s>