1. **State the problem:** Solve the quadratic equation $2x^2 + 5x + 3 = 0$. 2. **Formula used:** The quadratic formula is given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. **Identify coefficients:** Here, $a=2$, $b=5$, and $c=3$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 5^2 - 4 \times 2 \times 3 = 25 - 24 = 1$$ 5. **Apply the quadratic formula:** $$x = \frac{-5 \pm \sqrt{1}}{2 \times 2} = \frac{-5 \pm 1}{4}$$ 6. **Find the two solutions:** - For the plus sign: $$x = \frac{-5 + 1}{4} = \frac{-4}{4} = -1$$ - For the minus sign: $$x = \frac{-5 - 1}{4} = \frac{-6}{4} = \frac{\cancel{6}}{\cancel{4}} \times \frac{-1}{1} = -\frac{3}{2}$$ 7. **Final answer:** The solutions to the equation are $$x = -1 \quad \text{and} \quad x = -\frac{3}{2}$$