Question

Find the zeros of the quadratic equation y= 6x^2 + 13x + 5

Answer 1

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

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

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

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. Since the discriminant is positive, there will be two distinct real solutions. Now we apply the quadratic formula:

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

\[
\sqrt{49} = 7
\]

So the formula becomes:

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

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

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

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

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

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