1. **State the problem:** Solve the quadratic equation $$5r^2 + r - 2 = 0$$ for all real values of $r$. 2. **Recall the quadratic formula:** For an equation $$ax^2 + bx + c = 0$$, the solutions are given by $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=1$, and $c=-2$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 1^2 - 4 \times 5 \times (-2) = 1 + 40 = 41$$ Since $\Delta > 0$, there are two distinct real solutions. 4. **Apply the quadratic formula:** $$r = \frac{-1 \pm \sqrt{41}}{2 \times 5} = \frac{-1 \pm \sqrt{41}}{10}$$ 5. **Final answer:** The solutions in simplest form are $$r = \frac{-1 + \sqrt{41}}{10} \quad \text{and} \quad r = \frac{-1 - \sqrt{41}}{10}$$