To find where the quadratic equation \( y = 6x^2 + 13x + 5 \) equals zero, we use a formula called the quadratic formula. This formula looks like this:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation, \( a = 6 \), \( b = 13 \), and \( c = 5 \).
First, we calculate something called the discriminant, which helps us understand the solutions. We use the formula:
\[ b^2 - 4ac \]
So, we calculate:
- \( b^2 = 13^2 = 169 \)
- \( 4ac = 4 \times 6 \times 5 = 120 \)
Next, we subtract \( 120 \) from \( 169 \) to find the discriminant:
\[ 169 - 120 = 49 \]
Now, we can use the quadratic formula. We plug in our values:
\[ x = \frac{-13 \pm \sqrt{49}}{12} \]
Since the square root of \( 49 \) is \( 7 \), we can simplify it to:
\[ x = \frac{-13 \pm 7}{12} \]
This gives us two possible answers for \( x \):
- For the first solution, we calculate:
\[ x_1 = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2} \]
- For the second solution, we find:
\[ x_2 = \frac{-13 - 7}{12} = \frac{-20}{12} = -\frac{5}{3} \]
So, the solutions (or "zeros") of the equation \( y = 6x^2 + 13x + 5 \) are:
\[ x = -\frac{1}{2} \quad \text{and} \quad x = -\frac{5}{3} \]