To determine the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our function, we identify the coefficients \( a \), \( b \), and \( c \):
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the Discriminant
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \]
\[ 4ac = 4 \cdot 6 \cdot 5 = 120 \]
Now, compute the discriminant:
\[ b^2 - 4ac = 169 - 120 = 49 \]
Step 2: Apply the Quadratic Formula
Now that we have the discriminant, we can substitute \( a \), \( b \), and the discriminant into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Calculating \( \sqrt{49} \):
\[ \sqrt{49} = 7 \]
Now, substitute back into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
Step 3: Calculate the Two Possible Values of \( x \)
Now we calculate the two possible values for \( x \):
- Using the plus sign:
\[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- Using the minus sign:
\[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Final Result
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]
These values can also be presented as:
\[ \boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{-\frac{5}{3}} \]