1. **State the problem:** Solve the equation $$104 = 8y^2 + 8$$ for $y$. 2. **Isolate the term with $y^2$:** Subtract 8 from both sides: $$104 - 8 = 8y^2 + 8 - 8$$ $$96 = 8y^2$$ 3. **Divide both sides by 8 to solve for $y^2$:** $$\frac{96}{\cancel{8}} = \frac{8y^2}{\cancel{8}}$$ $$12 = y^2$$ 4. **Take the square root of both sides:** $$y = \pm \sqrt{12}$$ 5. **Simplify the square root:** $$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$$ 6. **Final answer:** $$y = \pm 2\sqrt{3}$$ This means $y$ can be either $2\sqrt{3}$ or $-2\sqrt{3}$.