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 \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). For the given function, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the Discriminant
First, we will calculate the discriminant \( D = b^2 - 4ac \):
\[ D = 13^2 - 4 \cdot 6 \cdot 5 \] \[ D = 169 - 120 \] \[ D = 49 \]
Step 2: Apply the Quadratic Formula
Now that we have \( D = 49 \), we can substitute \( a \), \( b \), and \( D \) into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Calculating the square root of 49:
\[ \sqrt{49} = 7 \]
Now substituting back into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
Step 3: Solve for the Two Solutions
Now we will calculate the two possible values for \( x \):
- First Solution \( x_1 \):
\[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- Second Solution \( x_2 \):
\[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]