1. **State the problem:** Find the solutions of the quadratic equation $$2x^2 + 3x = -8$$. 2. **Rewrite the equation in standard form:** Move all terms to one side: $$2x^2 + 3x + 8 = 0$$ 3. **Use the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 4. **Identify coefficients:** Here, $$a = 2$$, $$b = 3$$, and $$c = 8$$. 5. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 2 \times 8 = 9 - 64 = -55$$ 6. **Since the discriminant is negative, the solutions are complex:** $$x = \frac{-3 \pm \sqrt{-55}}{2 \times 2} = \frac{-3 \pm i\sqrt{55}}{4}$$ 7. **Final answer:** The solutions are $$x = \frac{-3 \pm i\sqrt{55}}{4}$$ This matches the option **-3 ± i√55 over 4**.