Answer
$9-x^2\leq0$
$(3-x)(3+x)\leq0$
$-(x-3)(x+3)\leq0$
$(x-3)(x+3)\geq0$
$x \leq-3$ and $3\leq x$
Work Step by Step
Step 1: Factorize inequality (difference of squares)
Step 2: Factorize $-1$ out of first Bracket (common factor)
Step 3: Multiply with $-1$ on both sides of the inequality
Step 4: Since the inequality was multiplied with a negative the direction of the inequality flips.
Step 5: The inequality function is greater than $0$ for $x \leq-3$ and $3\leq x$
