1. **State the problem:** Solve the quadratic equation $$8x^2 - 19x + 6 = 0$$ for $x$. 2. **Formula used:** To solve a quadratic equation of the form $$ax^2 + bx + c = 0$$, use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=8$, $b=-19$, and $c=6$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (-19)^2 - 4 \times 8 \times 6 = 361 - 192 = 169$$ Since $\Delta > 0$, there are two distinct real roots. 4. **Find the roots:** $$x = \frac{-(-19) \pm \sqrt{169}}{2 \times 8} = \frac{19 \pm 13}{16}$$ 5. **Evaluate each root:** - For the plus sign: $$x = \frac{19 + 13}{16} = \frac{32}{16} = 2$$ - For the minus sign: $$x = \frac{19 - 13}{16} = \frac{6}{16} = \frac{3}{8}$$ 6. **Final answer:** $$x = 2 \quad \text{or} \quad x = \frac{3}{8}$$ These correspond to option (a).