1. **Problem:** Find the real or non-real solutions to the equation $$x^2 + 0.5x - 14 = 0$$. 2. **Formula:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where the equation is in the form $$ax^2 + bx + c = 0$$. 3. **Identify coefficients:** Here, $$a = 1$$, $$b = 0.5$$, and $$c = -14$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (0.5)^2 - 4(1)(-14) = 0.25 + 56 = 56.25$$. 5. **Since $$\Delta > 0$$, there are two real solutions.** 6. **Apply the quadratic formula:** $$x = \frac{-0.5 \pm \sqrt{56.25}}{2(1)} = \frac{-0.5 \pm 7.5}{2}$$ 7. **Calculate each solution:** - $$x_1 = \frac{-0.5 + 7.5}{2} = \frac{7}{2} = 3.5$$ - $$x_2 = \frac{-0.5 - 7.5}{2} = \frac{-8}{2} = -4$$ **Final answer:** $$x = 3.5$$ or $$x = -4$$.