1. **State the problem:** Solve the quadratic equation by factorization: $$6y^2 - 149y - 120 = 0$$ 2. **Multiply the coefficient of $y^2$ and the constant term:** $$6 \times (-120) = -720$$ 3. **Find two numbers that multiply to $-720$ and add to $-149$:** These numbers are $-160$ and $+11$ because $$-160 \times 11 = -720$$ and $$-160 + 11 = -149$$ 4. **Rewrite the middle term using these two numbers:** $$6y^2 - 160y + 11y - 120 = 0$$ 5. **Group terms and factor each group:** $$ (6y^2 - 160y) + (11y - 120) = 0$$ $$ 2y(3y - 80) + 3(3y - 40) = 0$$ 6. **Factor out the common binomial:** $$ (3y - 40)(2y + 3) = 0$$ 7. **Set each factor equal to zero and solve for $y$:** $$3y - 40 = 0 \implies y = \frac{40}{3}$$ $$2y + 3 = 0 \implies y = -\frac{3}{2}$$ **Final answer:** $$y = \frac{40}{3} \text{ or } y = -\frac{3}{2}$$