Question

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

Answer 1

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

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

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

In our function, we have:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

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

\[
D = b^2 - 4ac = 13^2 - 4 \cdot 6 \cdot 5
\]

Calculating \( 13^2 \):

\[
13^2 = 169
\]

Calculating \( 4 \cdot 6 \cdot 5 \):

\[
4 \cdot 6 = 24 \quad \text{and} \quad 24 \cdot 5 = 120
\]

Now substituting these values back into the discriminant:

\[
D = 169 - 120 = 49
\]

### Step 2: Use the Quadratic Formula
Now that we have the discriminant, we can find the solutions using the quadratic formula. Since \( D = 49 \), we proceed as follows:

\[
x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]

Calculating \( \sqrt{49} \):

\[
\sqrt{49} = 7
\]

Now substituting back into the formula:

\[
x = \frac{-13 \pm 7}{12}
\]

### Step 3: Calculate the Two Solutions
Now we can calculate the two possible values for \( x \).

**First Solution:**

\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

**Second Solution:**

\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

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

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