1. **State the problem:** Solve the quadratic equation $q^2 + 19q = 10$ for $q$. 2. **Rewrite the equation:** Move all terms to one side to set the equation to zero: $$q^2 + 19q - 10 = 0$$ 3. **Identify coefficients:** For the quadratic equation $aq^2 + bq + c = 0$, here $a=1$, $b=19$, and $c=-10$. 4. **Use the quadratic formula:** $$q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 5. **Calculate the discriminant:** $$b^2 - 4ac = 19^2 - 4 \times 1 \times (-10) = 361 + 40 = 401$$ 6. **Substitute values into the formula:** $$q = \frac{-19 \pm \sqrt{401}}{2}$$ 7. **Simplify the expression:** The solutions are: $$q = \frac{-19 + \sqrt{401}}{2} \quad \text{or} \quad q = \frac{-19 - \sqrt{401}}{2}$$ 8. **Interpretation:** These are the two real roots of the quadratic equation. **Final answer:** $$q = \frac{-19 \pm \sqrt{401}}{2}$$