Solve for x:

$4x ^ { 1 \over 3} -2{x \over x^{2 \over 3}} = 7 + \sqrt {3x}$

This is the correct equation right?

Yes, I'll solve it now, I just wanted to make sure I had the right question.

Anyways, we have: $4x^{1 \over 3} - 2{x \over x^{2\over 3}} = 7 + \sqrt {3x}$

First, note that ${x \over {x ^ {2 \over 3}}} = {x^{3 \over 3} \over x^{2 \over 3}} = x^{1 \over 3}$

This means we have: $4x^{1 \over 3} - 2{x^{1\over 3}} = 7 + \sqrt {3x}$. 

Subtracting like terms gives us: $2x^{1 \over 3} = 7 + \sqrt{3x}$

To solve, let $y = x^{1 \over 3}$. This gives $2y = 7 + \sqrt{3y^3}$

We can convert this to $0=-3y^3+ 4y^2-28y+49$.

Solving for y gives us $y \approx 1.65543$, meaning $x = \sqrt[3]y \approx1.18296$. 

However, plugging this in doesn't work, meaning the solution is imaginary....

According to WA, its $\approx -0.161 + 3.137i$.

However, you should probably plug it in the original equation to see if it works.....