Solving $\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$ 
I need some tips with solving this algebraic equation
$$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$

I've tried subtracting: $$\frac{8x}{x+2}$$
and also setting $$\sqrt x = t$$
Where "t" is a substitution to make things simpler.
This is how it ends up:
$$x + 2 = 6\sqrt x - \frac{8x}{x+2}$$
or
$$t^2 + 2 = 6t - \frac{8t^2}{t^2+2}$$
Unfortunately, I couldn't find a way from here without getting polynomials of the fourth degree or equations with $$x\sqrt x$$

I'd just like to clarify that I'm not looking for the solution here. I'd just like it if I could have some pointers or tips on where to go from here, or even if I did something wrong.

Thanks in advance!!!
 A: Middle term splitting will always remain my favorite method for factorising polynomials!
\begin{align*}
&\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x\\
\iff &\dfrac{(x+2)^2+8x}{x+2}=6\sqrt x\\
\iff &(x+2)^2-6\sqrt x(x+2)+8x=0\\
\iff &(x+2)^2-4\sqrt x(x+2)-2\sqrt x(x+2)+8x=0\\
\iff &(x+2-4\sqrt x)(x+2-2\sqrt x)=0
\end{align*}
Now apply the quadratic formula.
A: Substitute $\sqrt x=t$ to get fourth degree polynomial equation which further can be factorized into two quadratics polynomials
$$\frac{t^4 + 12t^2 + 4}{t^2+2} = 6t$$
$$t^4-6t^3+12t^2-12t+4=0$$
$$(t^2-4t+2)(t^2-2t+2)=0$$
$$t^2-4t+2=0, \ \ \ \ t^2-2t+2=0$$
Solving above quadratic equations, we get
$$t=2\pm \sqrt2, \ \ \ t=1\pm i$$
A: Start with
$$\frac{x^2 + 12x + 4}{x+2} = 6\sqrt x$$
Then,
$$x^2+12x+4=6\sqrt{x}(x+2)$$
Square both sides,
$$ x^4 + 24 x^3 + 152 x^2 + 96 x + 16 = 36x^3+144x^2+144x$$
Therefore we want to find the roots of the equation
$$x^4-12x^3+8x^2 - 48x +16 =0$$
Although you could use the quartic formula it's a real pain. Wolfram Alpha finds
$$x=6 \pm 4\sqrt{2} \ ; x= \pm 2i.$$
