Is it possible to extract a variable from this equation? I have the following equation that I am trying to extract $x$ from but I'm stuck here:
$180\sqrt{\frac{x}{6}}+\frac{x}{48} = 2400$
Is it possible to get the value of x in this equation?
 A: HINT: write it in the form $$180\sqrt{\frac{x}{6}}=2400-\frac{x}{48}$$ and square it
A: Introducing 
$$
y = \sqrt{\frac{x}{6}}, \quad y^2 = \frac{x}{6}
$$
we get the equation
$$
180\, y + \frac{1}{8} y^2 = 2400
$$
This is a quadratic equation in $y$, and we get
$$
19200 = y^2 + 1440 \, y = (y + 720)^2 - 518400 \Rightarrow \\
y = \pm\sqrt{537600} - 720 \in \{ -1453.2,  13.212 \}
$$
We pick the positive solution $y = \sqrt{537600} - 720 = 13.212$.
Translating back:
$$
x = 6 \, y^2 = 6 \, \left( \sqrt{537600} - 720 \right)^2 = 1047.4
$$
A: HINT:
$$180\sqrt{\frac{x}{6}}+\frac{x}{48}=2400\Longleftrightarrow$$
$$30\sqrt{6}\sqrt{x}+\frac{x}{48}=2400\Longleftrightarrow$$
$$\frac{1}{48}\left(1440\sqrt{6}\sqrt{x}+x\right)=2400\Longleftrightarrow$$
$$1440\sqrt{6}\sqrt{x}+x=115200\Longleftrightarrow$$
$$1440\sqrt{6}\sqrt{x}=115200-x\Longleftrightarrow$$
$$12441600x=(115200-x)^2\Longleftrightarrow$$
$$12441600x=x^2-230400x+13271040000\Longleftrightarrow$$
$$12441600x-x^2+230400x-13271040000=0\Longleftrightarrow$$
$$-x^2+12672000x-13271040000=0\Longleftrightarrow$$
$$x^2-12672000x+13271040000=0$$
