Solve R 41C041

1. **State the problem:** We need to find the positive value of $r$ that satisfies the equation $$r(5 + 2r) - 5(r - 4) = 52$$. 2. **Expand the terms:** $$r(5 + 2r) = 5r + 2r^2$$ $$-5(r - 4) = -5r + 20$$ So the equation becomes: $$5r + 2r^2 - 5r + 20 = 52$$ 3. **Simplify the equation:** Combine like terms: $$5r - 5r = \cancel{5r} - \cancel{5r} = 0$$ So we have: $$2r^2 + 20 = 52$$ 4. **Isolate the quadratic term:** $$2r^2 = 52 - 20$$ $$2r^2 = 32$$ 5. **Divide both sides by 2:** $$\frac{2r^2}{\cancel{2}} = \frac{32}{\cancel{2}}$$ $$r^2 = 16$$ 6. **Solve for $r$:** $$r = \pm \sqrt{16}$$ $$r = \pm 4$$ 7. **Apply the condition that $r$ is positive:** Therefore, $$r = 4$$. **Final answer:** $$r = 4$$