1. The problem is to find the range of the quadratic function $$y=2(x-3)^2-4$$. 2. Identify the vertex form of the quadratic: $$y=a(x-h)^2+k$$ where the vertex is at $$(h,k)$$. 3. Here, $$a=2>0$$, $$h=3$$, $$k=-4$$, so the parabola opens upwards with vertex at $$(3,-4)$$. 4. Since the parabola opens upward, the lowest point of $$y$$ is at the vertex $$y=-4$$. 5. Therefore, the range is all values $$y$$ such that $$y \\geq -4$$. 6. Among options, this corresponds to $$A=\{y \,|\, y \\geq -4\}$$. Final answer: $$\boxed{A=\{y \,|\, y \\geq -4\}}$$