How do you find the zeroes of f(x) = x^4 -24x^2- 25?

${x}^{4} - 24 {x}^{2} - 25$
$= \left({x}^{2} - 25\right) \left({x}^{2} + 1\right)$
$= \left(x - 5\right) \left(x + 5\right) \left({x}^{2} + 1\right)$
${x}^{2} + 1 > 0$ for all $x \in \mathbb{R}$
So the only zeros are given by $x = - 5$ and $x = 5$.