6y^2+y-7=0

To solve the quadratic equation \(6y^2 + y - 7 = 0\), we can use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 6\), \(b = 1\), and \(c = -7\). Now, we can calculate the discriminant (\(D\)): \[ D = b^2 - 4ac = 1^2 - 4 \cdot 6 \cdot (-7) = 1 + 168 = 169 \] Since \(D\) is positive, there are two real solutions. Now we can substitute back into the quadratic formula: \[ y = \frac{-1 \pm \sqrt{169}}{2 \cdot 6} = \frac{-1 \pm 13}{12} \] Calculating the two possible values for \(y\): 1. \(y_1 = \frac{-1 + 13}{12} = \frac{12}{12} = 1\) 2. \(y_2 = \frac{-1 - 13}{12} = \frac{-14}{12} = -\frac{7}{6}\) Thus, the solutions to the equation \(6y^2 + y - 7 = 0\) are: \[ y = 1 \quad \text{or} \quad y = -\frac{7}{6} \]