Question 549327
If you changed variables to {{{y=x^2}}}
it would simplify into
{{{y^2-9y+20=0}}} which you would factor to get {{{(y-4)(y-5)=0}}}, so
{{{x^4-9x^2+20=0}}} factors into {{{(x^2-4)(x^2-5)=0}}}
It further factors into
{{{(x+2)(x-2)(x+sqrt(5))(x-sqrt(5))=0}}}
The 4 real solutions to that equation are:
{{{x=-2}}}, {{{x=2}}}, {{{x=-sqrt(5)}}}, and {{{x=sqrt(5)}}}