y = x\(^2\) - x - 12 = (x - 4)(x + 3) ∴ x = 4 or x = -3 Checking the cases for y \( \geq \) 0 We check values on the range x - 4 \(\geq\) 0; x + 3 \(\leq\) 0; x - 4 \(\leq\) 0 and x + 3 \(\geq\) 0 for the range which satisfies the inequality x\(^2\) - x - 12 \(\geq\) 0. We find that the inequality is satisfied on the range x \(\leq\) -3 and x \(\geq\) 4.