Question 357880
you need to be clearer as to what you want.

I am going to guess that you want to solve for x
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Let y={{{x^(1/3)}}}
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rewrite your equation
{{{30x^(2/3)-4x^(1/3)-2=0}}} to
 {{{30y^2-4y-2=0}}}
use quadratic equation to solve for y
a=30, b=-4, c=-2

y={{{(-(-4)-Sqrt((-4)^2-4*30*(-2)))/(2*30)}}} or y={{{(-(-4)+Sqrt((-4)^2-4*30*(-2)))/(2*30)}}}
y=-12/60=-1/5 or y=20/60=1/3
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since y={{{x^(1/3)}}} then

{{{x^(1/3)=-1/5}}} or {{{x^(1/3)=1/3}}}
X={{{(-1/5)^3=-1/125}}} or x={{{(1/3)^3=1/27}}}

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you should validate the answers by substituting the x values back in the original equation to verify if one or both work