To find 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: Identify coefficients
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 2: Calculate the discriminant
The discriminant (\( D \)) is given by:
\[ D = b^2 - 4ac \]
Substituting in the values of \( a \), \( b \), and \( c \):
\[ D = 13^2 - 4 \cdot 6 \cdot 5 \] \[ D = 169 - 120 \] \[ D = 49 \]
Step 3: Find the roots using the quadratic formula
Now that we have calculated the discriminant, we can substitute it back into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Calculating \( \sqrt{49} \):
\[ \sqrt{49} = 7 \]
Now substituting this value back into the equation:
\[ x = \frac{-13 \pm 7}{12} \]
Step 4: Calculate the two possible values for \( x \)
First root:
\[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
Second root:
\[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Step 5: State the zeros of the function
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]