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 our case, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the discriminant
First, we need to calculate the discriminant \( D \):
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = 13^2 - 4 \cdot 6 \cdot 5 \] \[ D = 169 - 120 \] \[ D = 49 \]
Step 2: Apply the quadratic formula
Now, we can substitute the values into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Since \( \sqrt{49} = 7 \), we have:
\[ x = \frac{-13 \pm 7}{12} \]
Step 3: Calculate the two possible values for \( x \)
Now we will find both values for \( x \):
-
For the plus case (+): \[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
-
For the minus case (−): \[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Step 4: Present the final results
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]
These are the points where the function intersects the x-axis.
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