Solve −x2+x+6≥0-x^2 + x + 6 \ge 0−x2+x+6≥0.A−2≤x≤3-2 \le x \le 3−2≤x≤3check_circleBx≤−2x \le -2x≤−2 or x≥3x \ge 3x≥3Cx≤−3x \le -3x≤−3 or x≥2x \ge 2x≥2D−3≤x≤2-3 \le x \le 2−3≤x≤2Explanation−(x2−x−6)≥0-(x^2 - x - 6) \ge 0−(x2−x−6)≥0, (x−3)(x+2)≤0(x-3)(x+2) \le 0(x−3)(x+2)≤0, so −2≤x≤3-2 \le x \le 3−2≤x≤3.