To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this case, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the Discriminant
First, we need to find 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
Since the discriminant is positive, we will have two distinct real zeros.
Now we can substitute \( a \), \( b \), and \( D \) into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Step 3: Calculate the Square Root
\[ \sqrt{49} = 7 \]
Step 4: Substitute and Simplify
Now substitute back into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
This gives us two solutions:
- First Solution:
\[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- Second Solution:
\[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Final Result
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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