1. **State the problem:** Solve the quadratic equation $$3x^2 - 20x - 13 = -6$$ by factoring. 2. **Rewrite the equation:** Move all terms to one side to set the equation equal to zero: $$3x^2 - 20x - 13 + 6 = 0$$ $$3x^2 - 20x - 7 = 0$$ 3. **Identify coefficients:** Here, $$a = 3$$, $$b = -20$$, and $$c = -7$$. 4. **Multiply $$a$$ and $$c$$:** $$3 \times (-7) = -21$$ 5. **Find two numbers that multiply to $$-21$$ and add to $$b = -20$$:** These numbers are $$-21$$ and $$1$$ because $$-21 \times 1 = -21$$ and $$-21 + 1 = -20$$. 6. **Rewrite the middle term using these numbers:** $$3x^2 - 21x + x - 7 = 0$$ 7. **Group terms:** $$(3x^2 - 21x) + (x - 7) = 0$$ 8. **Factor each group:** $$3x(x - 7) + 1(x - 7) = 0$$ 9. **Factor out the common binomial:** $$(3x + 1)(x - 7) = 0$$ 10. **Set each factor equal to zero and solve for $$x$$:** $$3x + 1 = 0 \Rightarrow 3x = -1 \Rightarrow x = \frac{-1}{3}$$ $$x - 7 = 0 \Rightarrow x = 7$$ **Final answer:** $$x = \frac{-1}{3} \text{ or } x = 7$$