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.

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.
