Determine x: Solving y = -2x² + 8x - 10 for Positive Values

To solve this problem, we'll check where the quadratic function $y = -2x^2 + 8x - 10$ is greater than zero.

The steps to solve are as follows:

• Step 1: Identify the direction of opening and calculate the vertex.
  For the quadratic function $y = ax^2 + bx + c$ with $a = -2$, $b = 8$, $c = -10$, since $a = -2 < 0$, the parabola opens downwards.
• Step 2: Find the vertex, which is the peak of the parabola.
  The x-coordinate of the vertex $x_v$ is given by the formula $x_v = -\frac{b}{2a} = -\frac{8}{2 \times (-2)} = 2$.
  Substitute $x = 2$ into the function to find the y-coordinate: $y = -2(2)^2 + 8(2) - 10 = -8 + 16 - 10 = -2$.
  The vertex is at $(2, -2)$. This indicates that the maximum value of the function is $-2$, which is less than zero.
• Step 3: Analyze the function's values.
  Since the maximum point $(2, -2)$ is less than zero and the parabola opens downwards, the function values are always less than or equal to the vertex's y-value, which is $-2$.
• Step 4: Conclusion.
  It is clear that the function is never positive at any point in its domain for real values of $x$.

Therefore, the solution to the problem is The function has no positive domain.