To determine the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we need to solve the equation \( 6x^2 + 13x + 5 = 0 \).
Step 1: Identify coefficients
The quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 2: Use the quadratic formula
The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Step 3: 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 4: Calculate the roots
Now we can substitute \( a \), \( b \), and the discriminant into the quadratic formula: \[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \] \[ x = \frac{-13 \pm 7}{12} \]
We will calculate the two possible values for \( x \):
First root: \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
Second root: \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are: \[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]
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