To determine 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 \( a = 6 \), \( b = 13 \), and \( c = 5 \).
Step 1: Calculate the discriminant (\( b^2 - 4ac \))
\[ b^2 - 4ac = 13^2 - 4(6)(5) \]
Calculating \( 13^2 \):
\[ 13^2 = 169 \]
Calculating \( 4 \times 6 \times 5 \):
\[ 4 \times 6 \times 5 = 120 \]
Now, plug these values into the discriminant:
\[ b^2 - 4ac = 169 - 120 = 49 \]
Step 2: Calculate the roots using the quadratic formula
Since the discriminant is positive, there are two real and distinct solutions.
Now substitute \( a \), \( b \), and the discriminant into the formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2(6)} \]
Since \( \sqrt{49} = 7 \), we can rewrite it as:
\[ x = \frac{-13 \pm 7}{12} \]
Now solve for the two possible values of \( x \):
- For the positive case:
\[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- For the negative case:
\[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Final Result:
The zero(s) of the quadratic function \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]