Question

Find the zeros of the quadratic equation y = 6x^2 + 13x + 5
Using 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, which is given by:

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

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

First, we need to calculate the discriminant \( b^2 - 4ac \):

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

Now we can substitute the values into the quadratic formula:

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

Since \( \sqrt{49} = 7 \), we have:

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

Now we calculate the two possible values for \( x \):

1. \( x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \)
2. \( x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \)

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

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