Solve for \( k \) \[ 2 k^{2}-13 k+11=0 \]

To solve the quadratic equation \( 2k^2 - 13k + 11 = 0 \), we can use the quadratic formula, which is given by: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this equation, \( a = 2 \), \( b = -13 \), and \( c = 11 \). 1. First, we calculate the discriminant \( b^2 - 4ac \): \[ b^2 = (-13)^2 = 169 \] \[ 4ac = 4 \cdot 2 \cdot 11 = 88 \] \[ b^2 - 4ac = 169 - 88 = 81 \] 2. Now we substitute \( b \) and the calculated discriminant into the quadratic formula: \[ k = \frac{-(-13) \pm \sqrt{81}}{2 \cdot 2} \] \[ k = \frac{13 \pm 9}{4} \] 3. This gives us two possible solutions: For the positive option: \[ k = \frac{13 + 9}{4} = \frac{22}{4} = \frac{11}{2} = 5.5 \] For the negative option: \[ k = \frac{13 - 9}{4} = \frac{4}{4} = 1 \] The solutions for \( k \) are \( k = 5.5 \) and \( k = 1 \).