To determine the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6 \), \( b = 13 \), and \( c = 5 \).
Step 1: Calculate the discriminant (\( b^2 - 4ac \))
First, we will compute the discriminant:
\[ b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5 \]
Calculating \( 13^2 \):
\[ 13^2 = 169 \]
Now calculating \( 4ac \):
\[ 4 \cdot 6 \cdot 5 = 120 \]
Now, substitute these values into the discriminant:
\[ b^2 - 4ac = 169 - 120 = 49 \]
Step 2: Apply the quadratic formula
Now, substituting the values into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Calculating \( \sqrt{49} \):
\[ \sqrt{49} = 7 \]
So the equation becomes:
\[ x = \frac{-13 \pm 7}{12} \]
Step 3: Solve for the two possible values of \( x \)
Now we will find the two values for \( x \):
- Using the positive sign:
\[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- Using the negative sign:
\[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Step 4: State the zeros
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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