Question 1208848
try something like  {{{t=x^(1/3)}}} and then you have   (t equals x to the one-third power)
{{{t^2+4t+2=0}}}


{{{t=(-4+- sqrt(4^2-4*2))/2}}}

{{{t=(-4+- sqrt(8))/2}}}

{{{t=(-4+- 2*sqrt(2))/2}}}

{{{t=-2+- sqrt(2)}}}


Two things to do now.

{{{x^(1/3)=-2-sqrt(2)}}}
The other possible x
{{{x^(1/3)=-2+sqrt(2)}}}


Cube both sides for each, to find x.