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