To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we need to solve the equation \( 6x^2 + 13x + 5 = 0 \). We can do this using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6 \), \( b = 13 \), and \( c = 5 \).
Step 1: Calculate the discriminant
First, we calculate the discriminant \( D = b^2 - 4ac \):
\[ D = 13^2 - 4(6)(5) \] \[ D = 169 - 120 \] \[ D = 49 \]
Step 2: Apply the quadratic formula
Now, we can apply the values into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2(6)} \]
Step 3: Simplify the square root
Since \( \sqrt{49} = 7 \), we can substitute this into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
Step 4: Calculate the two possible values of \( x \)
Now we will calculate the two possible solutions:
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First solution: \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
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Second solution: \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]