To find the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation in the standard form \( ax^2 + bx + c = 0 \).
For our equation:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)
Let's calculate the discriminant (\( b^2 - 4ac \)) first:
\[ b^2 = 12^2 = 144 \]
\[ 4ac = 4 \times (-3) \times 15 = -180 \]
Now, calculate the discriminant:
\[ b^2 - 4ac = 144 - (-180) = 144 + 180 = 324 \]
Since the discriminant is positive, we expect two real roots.
Now let's apply the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{324}}{2 \times -3} \]
First, we calculate \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
Now substitute back into the formula:
\[ x = \frac{-12 \pm 18}{-6} \]
This gives us two cases to solve for \( x \):
- Using the positive root:
\[ x = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \]
- Using the negative root:
\[ x = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \]
Thus, the zeros of the quadratic equation \( y = -3x^2 + 12x + 15 \) are:
\[ x = -1 \quad \text{and} \quad x = 5 \]