1. **State the problem:** Solve the quadratic equation $$5x^2 + 19x + 12 = 0$$ for $x$. 2. **Formula used:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=19$, and $c=12$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 19^2 - 4 \times 5 \times 12 = 361 - 240 = 121$$. 4. **Find the square root of the discriminant:** $$\sqrt{121} = 11$$. 5. **Apply the quadratic formula:** $$x = \frac{-19 \pm 11}{2 \times 5} = \frac{-19 \pm 11}{10}$$. 6. **Calculate the two roots:** - For the plus sign: $$x = \frac{-19 + 11}{10} = \frac{-8}{10} = -\frac{4}{5}$$. - For the minus sign: $$x = \frac{-19 - 11}{10} = \frac{-30}{10} = -3$$. 7. **Final answer:** The solutions are $$x = -3$$ and $$x = -\frac{4}{5}$$. These are the points where the parabola crosses the x-axis.