Solve the equation with radicals: $x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0$ Solve this equation :
$ x^2- x \sqrt[4]{2} (1+ \sqrt[4]{2} ) + \sqrt[4]{8} = 0 $
I've calculated $ \Delta $ and it is something like this: 
$ \sqrt[4]{4} + 4 \sqrt[4]{2} + 4 - 4 \sqrt[4]{8} $
I don't know what to do next.

Hint: Let $a=\sqrt[4]{2}$. So $a^3=\sqrt[4]{8}$. Now just use the quadratic formula
