To determine 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 \( a = 6 \), \( b = 13 \), and \( c = 5 \).
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \] \[ 4ac = 4 \times 6 \times 5 = 120 \]
So, the discriminant is:
\[ b^2 - 4ac = 169 - 120 = 49 \]
- Now, apply the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \times 6} \] \[ x = \frac{-13 \pm 7}{12} \]
- Calculate the two possible values for \( x \):
- For the positive case:
\[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- For the negative case:
\[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Thus, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]