1. **State the problem:** Solve the quadratic equation $$15b^2 - 8b - 3 = -4$$ for $b$ using the quadratic formula. 2. **Rewrite the equation in standard form:** Move all terms to one side to get zero on the other side: $$15b^2 - 8b - 3 + 4 = 0$$ $$15b^2 - 8b + 1 = 0$$ 3. **Identify coefficients:** Here, $a = 15$, $b = -8$, and $c = 1$. 4. **Recall the quadratic formula:** $$b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$$ where $A$, $B$, and $C$ are coefficients from the quadratic equation $Ax^2 + Bx + C = 0$. 5. **Substitute the values:** $$b = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 15 \times 1}}{2 \times 15}$$ 6. **Simplify inside the square root:** $$b = \frac{8 \pm \sqrt{64 - 60}}{30}$$ $$b = \frac{8 \pm \sqrt{4}}{30}$$ 7. **Calculate the square root:** $$b = \frac{8 \pm 2}{30}$$ 8. **Find the two possible solutions:** $$b = \frac{8 + 2}{30} = \frac{10}{30}$$ $$b = \frac{8 - 2}{30} = \frac{6}{30}$$ 9. **Simplify the fractions:** $$b = \frac{\cancel{10}}{\cancel{30}} = \frac{1}{3}$$ $$b = \frac{\cancel{6}}{\cancel{30}} = \frac{1}{5}$$ **Final answer:** $$b = \frac{1}{3} \text{ or } b = \frac{1}{5}$$