To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation in the form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the Discriminant
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 13^2 - 4 \times 6 \times 5 \] \[ = 169 - 120 \] \[ = 49 \]
Step 2: Apply the Quadratic Formula
Now we can substitute the values into the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-13 \pm \sqrt{49}}{2 \times 6} \] \[ = \frac{-13 \pm 7}{12} \]
Step 3: Calculate the Two Possible Values for \( x \)
Now, we will evaluate for both the plus and minus cases:
-
For the plus case: \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
-
For the minus case: \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Step 4: Write the Final Answer
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]
Thus, the zeros are \( x = -\frac{1}{2} \) and \( x = -\frac{5}{3} \).