1. The problem involves solving the equation $$y^2 = 128$$ for $y$. 2. To solve for $y$, we take the square root of both sides: $$y = \pm \sqrt{128}$$ 3. Simplify the square root by factoring 128: $$128 = 64 \times 2$$ 4. Using the property $$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$$, we get: $$y = \pm \sqrt{64 \times 2} = \pm \sqrt{64} \times \sqrt{2}$$ 5. Since $$\sqrt{64} = 8$$, this simplifies to: $$y = \pm 8\sqrt{2}$$ 6. Therefore, the solutions for $y$ are: $$y = 8\sqrt{2} \text{ or } y = -8\sqrt{2}$$ 7. The problem also states $y = 8\sqrt{2}$, which is the positive root. Final answer: $$y = 8\sqrt{2}$$