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:
- Calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \] \[ 4ac = 4 \cdot 6 \cdot 5 = 120 \] \[ b^2 - 4ac = 169 - 120 = 49 \]
- Calculate the square root of the discriminant:
\[ \sqrt{49} = 7 \]
- Now apply the quadratic formula:
\[ x = \frac{{-13 \pm 7}}{{2 \cdot 6}} = \frac{{-13 \pm 7}}{12} \]
- 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:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]