To determine the zero(s) of the quadratic function \( y = 6x^2 + 13x + 5 \), we will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a \), \( b \), and \( c \) are the coefficients from the standard form of the quadratic equation \( ax^2 + bx + c \).
In our case, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the discriminant
The first step is to compute the discriminant \( D = b^2 - 4ac \).
\[ D = 13^2 - 4 \times 6 \times 5 \] \[ D = 169 - 120 \] \[ D = 49 \]
Step 2: Apply the quadratic formula
Since the discriminant is positive (\( D = 49 \)), we will have two real and distinct solutions. Next, we substitute \( a \), \( b \), and \( D \) into the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \] \[ x = \frac{-13 \pm \sqrt{49}}{2 \times 6} \] \[ x = \frac{-13 \pm 7}{12} \]
Step 3: Calculate the two solutions
Now, we will calculate the two possible values for \( x \).
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For \( x_1 \): \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
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For \( x_2 \): \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]