i just used a substitution, y=x^2, so you end up with a quadratic y^2+y-20=0 which leads up to (y+5)(y-4)=0, y=-5 / y=4 the plug back in x^2, x^2 = -5, x^2 = 4, which leads to the same 4 solutions you got
I have simply rewritten the second term as 5X^2-4X^2 which yields X^4+5X^2-4X^2-20 Take X^2 as a common factor from the first 2 terms X^2(X^2+5)-4(X^2+5) (X^2-4)(X^2+5) Which yields X=+/-2 Or X=+√5 i X= -√5 i
