Answer
Solution set is $\{\frac{27}{8},-125\}$
Work Step by Step
$2x^{\frac{2}{3}}+7x^{\frac{1}{3}}-15=0$
This equation is equivalent to
$2(x^{\frac{1}{3}})^{2}+7x^{\frac{1}{3}}-15=0$
Substituting $y = x^{\frac{1}{3}}$,
$2y^{2}+7y-15=0$
By factoring,
$(2y-3)(y+5)=0$
$2y-3 = 0$ or $y+5=0$
$y=\frac{3}{2}$ or $y=-5$
Let $y=\frac{3}{2}$
Replace $y = x^{\frac{1}{3}}$
$x^{\frac{1}{3}}=\frac{3}{2}$
Cubing both sides of the equations,
$(x^{\frac{1}{3}})^{3}=(\frac{3}{2})^{3}$
$x=\frac{27}{8}$
Let $y=-5$
Replace $y = x^{\frac{1}{3}}$
$x^{\frac{1}{3}}=-5$
Cubing both sides of the equations,
$(x^{\frac{1}{3}})^{3}=(-5)^{3}$
$x=-125$
Solution set is $\{\frac{27}{8},-125\}$
