To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
In our function, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the Discriminant
First, we need to compute the discriminant \( D \), which is \( b^2 - 4ac \):
\[ D = b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5 \]
Calculating \( 13^2 \):
\[ 13^2 = 169 \]
Calculating \( 4 \cdot 6 \cdot 5 \):
\[ 4 \cdot 6 = 24 \quad \text{and} \quad 24 \cdot 5 = 120 \]
Now substituting these values back into the discriminant:
\[ D = 169 - 120 = 49 \]
Step 2: Use the Quadratic Formula
Now that we have the discriminant, we can find the solutions using the quadratic formula. Since \( D = 49 \), we proceed as follows:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Calculating \( \sqrt{49} \):
\[ \sqrt{49} = 7 \]
Now substituting back into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
Step 3: Calculate the Two Solutions
Now we can calculate the two possible values for \( x \).
First Solution:
\[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
Second Solution:
\[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Conclusion
Thus, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ \boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{-\frac{5}{3}} \]