Question

Find the zeros of the quadratic equation y=6x2+13x+5
. You may use any of the following methods: factoring, square roots, complete the square, or quadratic formula.

Answer 1

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

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

In this case, the coefficients are:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)

Now we can calculate the values step by step:

1. **Calculate the discriminant** \( b^2 - 4ac \):

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

2. **Calculate the square root of the discriminant**:

\[
\sqrt{49} = 7
\]

3. **Now apply the quadratic formula**:

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

4. **Calculate the two possible values for \( x \)**:

- For the plus sign:
\[
x_1 = \frac{{-13 + 7}}{{12}} = \frac{{-6}}{{12}} = -\frac{1}{2}
\]

- For the minus sign:
\[
x_2 = \frac{{-13 - 7}}{{12}} = \frac{{-20}}{{12}} = -\frac{5}{3}
\]

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

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