Solve for x: 3.1 x2-2x-3>0, 3.2 x2+x-6<0, 3.3 x2-2x≥0, 3.4 x2+x-12≤0, 3.5 2-x-x2>0, 3.6 10+x≤2x2, 3.7 x2>4

3.1 $x^2 - 2x - 3 > 0$ Step 1: Find the roots of the quadratic equation $x^2 - 2x - 3 = 0$. $(x-3)(x+1) = 0$ $x = 3 \quad or \quad x = -1$ Step 2: Sketch the parabola $y = x^2 - 2x - 3$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $x^2 - 2x - 3 > 0$ means we are looking for the values of $x$ where the parabola is above the x-axis. Solution: $\boxed{x < -1 \quad or \quad x > 3}$

3.2 $x^2 + x - 6 < 0$ Step 1: Find the roots of the quadratic equation $x^2 + x - 6 = 0$. $(x+3)(x-2) = 0$ $x = -3 \quad or \quad x = 2$ Step 2: Sketch the parabola $y = x^2 + x - 6$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $x^2 + x - 6 < 0$ means we are looking for the values of $x$ where the parabola is below the x-axis. Solution: $\boxed{-3 < x < 2}$

3.3 $x^2 - 2x \ge 0$ Step 1: Find the roots of the quadratic equation $x^2 - 2x = 0$. $x(x-2) = 0$ $x = 0 \quad or \quad x = 2$ Step 2: Sketch the parabola $y = x^2 - 2x$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $x^2 - 2x \ge 0$ means we are looking for the values of $x$ where the parabola is above or on the x-axis. Solution: $\boxed{x \le 0 \quad or \quad x \ge 2}$

3.4 $x^2 + x - 12 \le 0$ Step 1: Find the roots of the quadratic equation $x^2 + x - 12 = 0$. $(x+4)(x-3) = 0$ $x = -4 \quad or \quad x = 3$ Step 2: Sketch the parabola $y = x^2 + x - 12$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $x^2 + x - 12 \le 0$ means we are looking for the values of $x$ where the parabola is below or on the x-axis. Solution: $\boxed{-4 \le x \le 3}$

3.5 $2 - x - x^2 > 0$ Step 1: Rewrite the inequality by multiplying by -1 and reversing the sign. $-x^2 - x + 2 > 0$ $x^2 + x - 2 < 0$ Step 2: Find the roots of the quadratic equation $x^2 + x - 2 = 0$. $(x+2)(x-1) = 0$ $x = -2 \quad or \quad x = 1$ Step 3: Sketch the parabola $y = x^2 + x - 2$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $x^2 + x - 2 < 0$ means we are looking for the values of $x$ where the parabola is below the x-axis. Solution: $\boxed{-2 < x < 1}$

3.6 $10 + x \le 2x^2$ Step 1: Rearrange the inequality to have 0 on one side. $0 \le 2x^2 - x - 10$ $2x^2 - x - 10 \ge 0$ Step 2: Find the roots of the quadratic equation $2x^2 - x - 10 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-10)}}{2(2)}$ $x = \frac{1 \pm \sqrt{1 + 80}}{4}$ $x = \frac{1 \pm \sqrt{81}}{4}$ $x = \frac{1 \pm 9}{4}$ $x_1 = \frac{1 + 9}{4} = \frac{10}{4} = \frac{5}{2}$ $x_2 = \frac{1 - 9}{4} = \frac{-8}{4} = -2$ Step 3: Sketch the parabola $y = 2x^2 - x - 10$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $2x^2 - x - 10 \ge 0$ means we are looking for the values of $x$ where the parabola is above or on the x-axis. Solution: $\boxed{x \le -2 \quad or \quad x \ge \frac{5}{2}}$

3.7 $x^2 > 4$ Step 1: Rearrange the inequality to have 0 on one side. $x^2 - 4 > 0$ Step 2: Find the roots of the quadratic equation $x^2 - 4 = 0$. $(x-2)(x+2) = 0$ $x = 2 \quad or \quad x = -2$ Step 3: Sketch the parabola $y = x^2 - 4$. Since the leading coefficient is positive, the parabola opens upwards. The inequality $x^2 - 4 > 0$ means we are looking for the values of $x$ where the parabola is above the x-axis. Solution: $\boxed{x < -2 \quad or \quad x > 2}$