1. **State the problem:** Solve the equation $$2y(y + 6) = 14$$ and find the solution set. 2. **Expand the left side:** Use the distributive property: $$2y(y + 6) = 2y \cdot y + 2y \cdot 6 = 2y^2 + 12y$$ So the equation becomes: $$2y^2 + 12y = 14$$ 3. **Bring all terms to one side:** Subtract 14 from both sides: $$2y^2 + 12y - 14 = 0$$ 4. **Simplify the equation:** Divide the entire equation by 2 to make coefficients smaller: $$\cancel{2}y^2 + \cancel{2} \cdot 6 y - \cancel{2} \cdot 7 = 0 \implies y^2 + 6y - 7 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=6$, and $c=-7$. 6. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-7) = 36 + 28 = 64$$ 7. **Find the roots:** $$y = \frac{-6 \pm \sqrt{64}}{2 \cdot 1} = \frac{-6 \pm 8}{2}$$ 8. **Calculate each solution:** - For the plus sign: $$y = \frac{-6 + 8}{2} = \frac{2}{2} = 1$$ - For the minus sign: $$y = \frac{-6 - 8}{2} = \frac{-14}{2} = -7$$ 9. **Solution set:** $$\{1, -7\}$$ 10. **Check the solutions:** - For $y=1$: $$2 \cdot 1 (1 + 6) = 2 \cdot 1 \cdot 7 = 14$$ (True) - For $y=-7$: $$2 \cdot (-7) (-7 + 6) = 2 \cdot (-7) \cdot (-1) = 14$$ (True) Both solutions satisfy the equation. **Final answer:** The solution set is $$\boxed{\{1, -7\}}$$.