Solve for \( x \) in the following equation by the quadratic formula method \( (3-5 x)^{2}=(2 x+2)^{2} \)

To solve the equation \( (3-5x)^{2}=(2x+2)^{2} \), we start by expanding both sides. Expanding the left side: \[ (3 - 5x)^{2} = 9 - 30x + 25x^{2} \] Expanding the right side: \[ (2x + 2)^{2} = 4x^{2} + 8x + 4 \] Now, we set both expanded forms equal to each other: \[ 9 - 30x + 25x^{2} = 4x^{2} + 8x + 4 \] Next, we move everything to one side of the equation to set it to 0: \[ 25x^{2} - 4x^{2} - 30x - 8x + 9 - 4 = 0 \] \[ 21x^{2} - 38x + 5 = 0 \] Now, we apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 21, b = -38, c = 5 \). Calculating the discriminant: \[ b^{2} - 4ac = (-38)^{2} - 4(21)(5) = 1444 - 420 = 1024 \] Now substituting into the quadratic formula: \[ x = \frac{-(-38) \pm \sqrt{1024}}{2(21)} = \frac{38 \pm 32}{42} \] This gives us two potential solutions: \[ x_1 = \frac{70}{42} = \frac{5}{3} \] \[ x_2 = \frac{6}{42} = \frac{1}{7} \] Thus, the solutions for \( x \) are \( \frac{5}{3} \) and \( \frac{1}{7} \).