1. **State the problem:** Solve the quadratic equation by factoring: $$4x^2 - 19x - 5 = 0$$ 2. **Recall the factoring method:** For a quadratic equation $$ax^2 + bx + c = 0$$, we look for two numbers that multiply to $$a \times c$$ and add to $$b$$. 3. **Calculate product and sum:** Here, $$a=4$$, $$b=-19$$, and $$c=-5$$. Calculate product: $$4 \times (-5) = -20$$. We need two numbers that multiply to $$-20$$ and add to $$-19$$. 4. **Find the two numbers:** The numbers are $$-20$$ and $$1$$ because $$-20 \times 1 = -20$$ and $$-20 + 1 = -19$$. 5. **Rewrite the middle term:** Rewrite $$-19x$$ as $$-20x + x$$: $$4x^2 - 20x + x - 5 = 0$$ 6. **Group terms:** $$(4x^2 - 20x) + (x - 5) = 0$$ 7. **Factor each group:** $$4x(x - 5) + 1(x - 5) = 0$$ 8. **Factor out common binomial:** $$(4x + 1)(x - 5) = 0$$ 9. **Set each factor equal to zero:** $$4x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{4}$$ $$x - 5 = 0 \quad \Rightarrow \quad x = 5$$ **Final answer:** $$x = -\frac{1}{4}$$ or $$x = 5$$