6y^2+y-7=0

To solve the quadratic equation \(6y^2 + y - 7 = 0\), we can use the quadratic formula:

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

Here, the coefficients are:

• \(a = 6\)
• \(b = 1\)
• \(c = -7\)

1. Calculate the discriminant \(b^2 - 4ac\):

\[ b^2 - 4ac = 1^2 - 4 \times 6 \times (-7) = 1 + 168 = 169 \]

2. Now, use the quadratic formula:

\[ y = \frac{-1 \pm \sqrt{169}}{2 \times 6} \]

3. Calculate \(\sqrt{169}\):

\[ \sqrt{169} = 13 \]

4. Substitute back into the formula:

\[ y = \frac{-1 \pm 13}{12} \]

5. This results in two possible values for \(y\):

\[ y_1 = \frac{-1 + 13}{12} = \frac{12}{12} = 1 \]

\[ y_2 = \frac{-1 - 13}{12} = \frac{-14}{12} = -\frac{7}{6} \]

Thus, the solutions to the equation \(6y^2 + y - 7 = 0\) are:

\[ y = 1 \quad \text{and} \quad y = -\frac{7}{6} \]