Question

Determine the zero or zeros of the quadratic function y = 6x^2 + 13x + 5. Show all the steps, but make them simple.

Answer 1

To find the zeros of the quadratic function \( y = 6x^2 + 13x + 5 \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).

From our quadratic function:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

### Step 1: Calculate the Discriminant
First, we need to calculate the discriminant \( b^2 - 4ac \):

\[
b^2 = 13^2 = 169
\]

\[
4ac = 4 \cdot 6 \cdot 5 = 120
\]

Now, substitute these values into the discriminant formula:

\[
b^2 - 4ac = 169 - 120 = 49
\]

### Step 2: Calculate the Square Root of the Discriminant
Next, we find the square root of the discriminant:

\[
\sqrt{49} = 7
\]

### Step 3: Apply the Quadratic Formula
Now, we can substitute \( a \), \( b \), and the square root of the discriminant into the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-13 \pm 7}{2 \cdot 6}
\]

### Step 4: Calculate the Two Possible Values for \( x \)

1. **First Solution**:

\[
x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. **Second Solution**:

\[
x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

### Conclusion
The zeros of the quadratic function \( y = 6x^2 + 13x + 5 \) are:

\[
x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3}
\]