Solve the equation: \( 2(2 a-1)^{2}-3(2 a-1)-5=0 \)

To solve the equation \( 2(2 a-1)^{2}-3(2 a-1)-5=0 \), let's use substitution. Let \( x = 2a - 1 \). Then the equation becomes: \[ 2x^2 - 3x - 5 = 0. \] We can now apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -3 \), and \( c = -5 \). Calculating the discriminant: \[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49. \] Now, substituting into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{49}}{2 \cdot 2} = \frac{3 \pm 7}{4}. \] This results in: \[ x_1 = \frac{10}{4} = \frac{5}{2}, \quad x_2 = \frac{-4}{4} = -1. \] Now, substituting back for \( x \): 1. For \( x_1 = \frac{5}{2} \): \[ 2a - 1 = \frac{5}{2} \implies 2a = \frac{7}{2} \implies a = \frac{7}{4}. \] 2. For \( x_2 = -1 \): \[ 2a - 1 = -1 \implies 2a = 0 \implies a = 0. \] Thus, the solutions to the equation are: \[ a = \frac{7}{4} \quad \text{and} \quad a = 0. \]