To locate the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we need to set the equation equal to zero and solve for \( x \):
\[ -3x^2 + 12x + 15 = 0 \]
We can simplify this equation by dividing all terms by -3:
\[ x^2 - 4x - 5 = 0 \]
Next, we can factor the quadratic equation:
\[ (x - 5)(x + 1) = 0 \]
Setting each factor to zero gives us:
- \( x - 5 = 0 \) → \( x = 5 \)
- \( x + 1 = 0 \) → \( x = -1 \)
The zeros of the function correspond to the points where the function intersects the x-axis. Thus, we have the zeros at:
\[ (5, 0) \quad \text{and} \quad (-1, 0) \]
Therefore, the correct answer is:
D) (-1, 0) (5, 0)