1. **State the problem:** Find the zeros (roots) of the quadratic equation $$5x^2 + 19x + 12 = 0$$. 2. **Formula and approach:** To find the roots, we can factor the quadratic if possible or use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=19$, and $c=12$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 19^2 - 4 \times 5 \times 12 = 361 - 240 = 121$$ Since $\Delta > 0$, there are two real roots. 4. **Find the roots using the quadratic formula:** $$x = \frac{-19 \pm \sqrt{121}}{2 \times 5} = \frac{-19 \pm 11}{10}$$ 5. **Calculate each root:** - For the plus sign: $$x = \frac{-19 + 11}{10} = \frac{-8}{10} = -\frac{4}{5}$$ - For the minus sign: $$x = \frac{-19 - 11}{10} = \frac{-30}{10} = -3$$ 6. **Write the factored form:** Using the roots $x = -\frac{4}{5}$ and $x = -3$, the factored form is: $$5x^2 + 19x + 12 = 5(x + 3)\left(x + \frac{4}{5}\right)$$ Multiply inside the second factor by 5 to clear the fraction: $$= (x + 3)(5x + 4)$$ **Final answer:** The zeros are $x = -3$ and $x = -\frac{4}{5}$, and the factored form is: $$ (x + 3)(5x + 4) = 0 $$