1. **State the problem:** Solve the quadratic equation $$10x^2 + x + 3 = 6$$ using the quadratic formula. 2. **Rewrite the equation in standard form:** Move all terms to one side: $$10x^2 + x + 3 - 6 = 0$$ $$10x^2 + x - 3 = 0$$ 3. **Identify coefficients:** Here, $$a = 10$$, $$b = 1$$, and $$c = -3$$. 4. **Recall the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 5. **Calculate the discriminant:** $$b^2 - 4ac = 1^2 - 4 \times 10 \times (-3) = 1 + 120 = 121$$ 6. **Substitute values into the formula:** $$x = \frac{-1 \pm \sqrt{121}}{2 \times 10} = \frac{-1 \pm 11}{20}$$ 7. **Find the two solutions:** - For the plus sign: $$x = \frac{-1 + 11}{20} = \frac{10}{20} = \frac{1}{2}$$ - For the minus sign: $$x = \frac{-1 - 11}{20} = \frac{-12}{20} = \frac{\cancel{\!12}}{\cancel{\!20}} = \frac{-3}{5}$$ 8. **Final answer:** $$x = \frac{1}{2} \quad \text{or} \quad x = -\frac{3}{5}$$