1. **State the problem:** Solve the quadratic equation $$2r^2 + r - 3 = 0$$ by factoring. 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, $$a = 2$$, $$b = 1$$, and $$c = -3$$. Calculate $$a \times c = 2 \times (-3) = -6$$. We need two numbers that multiply to $$-6$$ and add to $$1$$. 4. **Find the numbers:** The numbers are $$3$$ and $$-2$$ because $$3 \times (-2) = -6$$ and $$3 + (-2) = 1$$. 5. **Rewrite the middle term:** Rewrite $$r$$ as $$3r - 2r$$: $$2r^2 + 3r - 2r - 3 = 0$$ 6. **Group terms:** $$(2r^2 + 3r) - (2r + 3) = 0$$ 7. **Factor each group:** $$r(2r + 3) - 1(2r + 3) = 0$$ 8. **Factor out the common binomial:** $$(2r + 3)(r - 1) = 0$$ 9. **Set each factor equal to zero:** $$2r + 3 = 0 \quad \text{or} \quad r - 1 = 0$$ 10. **Solve for $$r$$:** $$2r + 3 = 0 \Rightarrow 2r = -3 \Rightarrow r = \frac{-3}{2}$$ $$r - 1 = 0 \Rightarrow r = 1$$ **Final answer:** The solutions are $$r = 1$$ and $$r = -\frac{3}{2}$$. Note: None of the provided options exactly match these roots, but the closest integer roots are not correct. The exact roots are $$1$$ and $$-1.5$$.