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 \( c = 5 \).
Step 1: Calculate the discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \] \[ 4ac = 4 \cdot 6 \cdot 5 = 120 \] \[ b^2 - 4ac = 169 - 120 = 49 \]
Step 2: Use the quadratic formula
Now that we have the discriminant, we can substitute into the quadratic formula:
\[ x = \frac{{-13 \pm \sqrt{49}}}{2 \cdot 6} \]
Since \( \sqrt{49} = 7 \), we have:
\[ x = \frac{{-13 \pm 7}}{12} \]
Step 3: Calculate the two possible values for \( x \)
Now we can solve for both possible values of \( x \):
- For the positive case:
\[ x = \frac{{-13 + 7}}{12} = \frac{{-6}}{12} = -\frac{1}{2} \]
- For the negative case:
\[ x = \frac{{-13 - 7}}{12} = \frac{{-20}}{12} = -\frac{20}{12} = -\frac{5}{3} \]
Final Result:
The zeros of the function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]