To find the zero(s) of the quadratic function \( y = 6x^2 + 13x + 5 \), we need to solve the equation \( 6x^2 + 13x + 5 = 0 \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c \). For our function:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
Step 1: Calculate the Discriminant
The first step is to compute the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \] \[ 4ac = 4 \cdot 6 \cdot 5 = 120 \] \[ b^2 - 4ac = 169 - 120 = 49 \]
Step 2: Apply the Quadratic Formula
Now, substitute the values of \( a \), \( b \), and the discriminant into the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \] \[ \sqrt{49} = 7 \] Now substituting back into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
Step 3: Find the Two Solutions
Now we will calculate the two possible values for \( x \):
- Using the positive square root:
\[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- Using the negative square root:
\[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Conclusion
Thus, the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are
\[ \boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{-\frac{5}{3}}. \]
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