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.

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.
