1. **State the problem:** Find the real solution(s) to the quadratic equation $$4x^2 + 8x + 3 = 0$$. 2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 3. **Identify coefficients:** Here, $$a=4$$, $$b=8$$, and $$c=3$$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 8^2 - 4 \times 4 \times 3 = 64 - 48 = 16$$. 5. **Check the discriminant:** Since $$\Delta = 16 > 0$$, there are two distinct real solutions. 6. **Apply the quadratic formula:** $$x = \frac{-8 \pm \sqrt{16}}{2 \times 4} = \frac{-8 \pm 4}{8}$$. 7. **Find each solution:** - For the plus sign: $$x = \frac{-8 + 4}{8} = \frac{-4}{8} = -\frac{1}{2}$$. - For the minus sign: $$x = \frac{-8 - 4}{8} = \frac{-12}{8} = -\frac{3}{2}$$. **Final answer:** The real solutions are $$x = -\frac{1}{2}$$ and $$x = -\frac{3}{2}$$.