To find the zeros of the quadratic function \( y = -3x^2 + 12x + 15 \), we can use the quadratic formula.
The general form of a quadratic equation is:
\[ ax^2 + bx + c = 0 \]
In this case, we have:
- \( a = -3 \)
- \( b = 12 \)
- \( c = 15 \)
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 12^2 = 144 \] \[ 4ac = 4 \times -3 \times 15 = -180 \] \[ b^2 - 4ac = 144 - (-180) = 144 + 180 = 324 \]
Now, we substitute the values into the quadratic formula:
\[ x = \frac{-12 \pm \sqrt{324}}{2 \times -3} \]
Calculating \( \sqrt{324} \):
\[ \sqrt{324} = 18 \]
Now substituting back into the formula:
\[ x = \frac{-12 \pm 18}{-6} \]
This gives us two potential zeros:
-
For the positive case: \[ x_1 = \frac{-12 + 18}{-6} = \frac{6}{-6} = -1 \]
-
For the negative case: \[ x_2 = \frac{-12 - 18}{-6} = \frac{-30}{-6} = 5 \]
Thus, the zeros of the function \( y = -3x^2 + 12x + 15 \) are:
\[ x = -1 \quad \text{and} \quad x = 5 \]
In summary, the roots (or zeros) of the equation are \( x = -1 \) and \( x = 5 \).