To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula, which is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our quadratic function, the coefficients are:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Now, we first need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \] \[ 4ac = 4 \times 6 \times 5 = 120 \] \[ b^2 - 4ac = 169 - 120 = 49 \]
Next, we compute the zeros using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-13 \pm \sqrt{49}}{2 \times 6} \] \[ x = \frac{-13 \pm 7}{12} \]
Now we can calculate the two possible values for \( x \):
-
Using \( -13 + 7 \): \[ x = \frac{-6}{12} = -\frac{1}{2} \]
-
Using \( -13 - 7 \): \[ x = \frac{-20}{12} = -\frac{5}{3} \]
The zeros of the function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]
Thus, the final answer is:
\[ \boxed{-\frac{1}{2}, -\frac{5}{3}} \]
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