Question 837887
The solution is very long a little tough...but I will summarize...


square both sides to get

{{{(x^2-5)^2=x+5}}}

Simplify and rewrite as

{{{P=x^4-10x^2-x+20}}}

We wish to write this as a product of square polynomials, so after some careful thought and experimentation we use


{{{P=(x^2-a)^2-(bx+c)^2}}}


When we expand this version and compare coefficients we can determine that


1) {{{-2a-b^2=-10}}}
2) {{{-2bc=-1}}}
3) {{{a^2-c^2=20}}}


This nonlinear can be solved by substitution. It has a rational solution of 

{{{a=9/2}}}
{{{b=1}}}
{{{c=1/2}}}


therefore we can write:


{{{P=(x^2-9/2)^2-(x+1/2)^2}}}

and that means

{{{x^4-10x^2-x+20=(x^2-9/2)^2-(x+1/2)^2}}}


so that

{{{x^4-10x^2-x+20=(x^2+x-5)(x^2+x-4)}}}


and we know there are only two solutions so...
{{{x=(-1-sqrt(17))/2}}}

{{{x=(1+sqrt(21))/2}}}