1. **State the problem:** Solve the quadratic equation $2x^2 - 13x + 20 = 0$. 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, $a = 2$, $b = -13$, and $c = 20$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-13)^2 - 4 \times 2 \times 20 = 169 - 160 = 9$$ Since $\Delta > 0$, there are two real and distinct solutions. 5. **Apply the quadratic formula:** $$x = \frac{-(-13) \pm \sqrt{9}}{2 \times 2} = \frac{13 \pm 3}{4}$$ 6. **Find the two solutions:** - For the plus sign: $$x = \frac{13 + 3}{4} = \frac{16}{4} = 4$$ - For the minus sign: $$x = \frac{13 - 3}{4} = \frac{10}{4} = 2.5$$ **Final answer:** The solutions are $x = 4$ and $x = 2.5$.