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, the coefficients are:
- \( a = 6 \)
- \( b = 13 \)
- \( c = 5 \)
First, we calculate the discriminant \( b^2 - 4ac \):
\[ b^2 = 13^2 = 169 \]
\[ 4ac = 4 \cdot 6 \cdot 5 = 120 \]
Now, we find the discriminant:
\[ b^2 - 4ac = 169 - 120 = 49 \]
Since the discriminant is positive, there will be two real solutions. Now we apply the quadratic formula:
\[ x = \frac{-13 \pm \sqrt{49}}{2 \cdot 6} \]
Calculating the square root:
\[ \sqrt{49} = 7 \]
Now we substitute back into the formula:
\[ x = \frac{-13 \pm 7}{12} \]
This gives us two solutions:
-
For the positive case: \[ x = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
-
For the negative case: \[ x = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
Thus, the zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2}, \quad x = -\frac{5}{3} \]