Question

Determine the zero(s) of the quadratic function y=6x^2+13x+5
. Show all steps.
write it small please

Answer 1

shorter please

Answer 2

shorter please

Answer 3

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 \( a = 6 \), \( b = 13 \), and \( c = 5 \).

**Step 1: Calculate the discriminant \( D \)**

\[
D = b^2 - 4ac
\]
\[
D = 13^2 - 4 \cdot 6 \cdot 5
\]
\[
D = 169 - 120
\]
\[
D = 49
\]

**Step 2: Apply the quadratic formula**

Since \( D = 49 \) (which is positive), there will be two real roots.

\[
x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6}
\]
\[
x = \frac{-13 \pm 7}{12}
\]

**Step 3: Calculate the two solutions**

1. **First root:**
\[
x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
\]

2. **Second root:**
\[
x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3}
\]

**Final solution:**

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

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

Answer 4

To find the zeros of \( y = 6x^2 + 13x + 5 \):

1. **Use the quadratic formula**:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 6 \), \( b = 13 \), \( c = 5 \).

2. **Calculate discriminant**:
\[
D = 13^2 - 4(6)(5) = 169 - 120 = 49
\]

3. **Solve for \( x \)**:
\[
x = \frac{-13 \pm 7}{12}
\]
- First root: \( x_1 = \frac{-6}{12} = -\frac{1}{2} \)
- Second root: \( x_2 = \frac{-20}{12} = -\frac{5}{3} \)

**Zeros**:
\[
x = -\frac{1}{2}, \quad x = -\frac{5}{3}
\]

Answer 5

To find the zeros of \( y = 6x^2 + 13x + 5 \):

1. **Discriminant**:
\[
D = 13^2 - 4(6)(5) = 49
\]

2. **Quadratic formula**:
\[
x = \frac{-13 \pm 7}{12}
\]

3. **Roots**:
\[
x_1 = -\frac{1}{2}, \quad x_2 = -\frac{5}{3}
\]

**Zeros**: \( x = -\frac{1}{2}, -\frac{5}{3} \)