Answer
$x=\pm5$
Work Step by Step
The zeros of the function are the $x$-intercepts of this function. The $x$-intercepts can be found by equating function expression to $0$ and solving for $x$:
$$f(x)=-x^2+25$$
$$-x^2+25=0$$
$$-x^2=-25$$
$$x^2=25$$
$$x=\pm\sqrt{25}=\pm5$$