Question 353894
5x^2/3 + 2x^4/3 -13 =0 
My first inclination was to rearrange the terms to... 
2x^4/3 + 5x^2/3 -13 =0 
This puts it in the right order for substitution / y= x^2/3 
Thus... 2y^2 + 5y -13 =0
This is where I'm stuck. I don't know how to unfold the answer with the unusual powers. 
Thanks for your help.
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Solve your equation for y:
*[invoke solve_quadratic_equation 2,5,-13]
y = -5/4 ± sqrt(129)/4
x^(2/3) = -5/4 ± sqrt(129)/4
It's messy, but it's just arithmetic from here.
Cube both sides:
x^(2/3) = -5/4 + sqrt(129)/4
x^2 = (1/16)*(-515 + 51sqrt(129))
{{{x = + (1/4)*sqrt(-515 + 51sqrt(129))}}}
{{{x = - (1/4)*sqrt(-515 + 51sqrt(129))}}}

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Can you do rest?