Find the real roots of the equation x^4-5x^2-36=0

Treat x^4 as the square of p:  x^4 = p^2.

Then x^4 - 5x^2 - 36 = 0 becomes p^2 - 5p - 36 = 0.

This factors nicely, to (p-9)(p+4) = 0.  Then p = 9 and p = -4.

Equating 9 and x^2, we find that x=3 or x=-3.
Equating -4 and x^2, we see that there's no real solution.

Show that both x=3 and x=-3 are real roots of x^4 - 5x^2 - 36 = 0.