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

To solve the equation \( (3-5x)^{2} = (2x+2)^{2} \) using the quadratic formula, we first expand both sides: 1. Expand \( (3 - 5x)^{2} \): \[ (3 - 5x)(3 - 5x) = 9 - 30x + 25x^{2} = 25x^{2} - 30x + 9 \] 2. Expand \( (2x + 2)^{2} \): \[ (2x + 2)(2x + 2) = 4x^{2} + 8x + 4 \] Now that we have expanded both sides, we can set the equation: \[ 25x^{2} - 30x + 9 = 4x^{2} + 8x + 4 \] Next, we move all terms to one side of the equation: \[ 25x^{2} - 4x^{2} - 30x - 8x + 9 - 4 = 0 \] This simplifies to: \[ 21x^{2} - 38x + 5 = 0 \] Now, we apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 21 \), \( b = -38 \), and \( c = 5 \). First, we need to calculate the discriminant: \[ b^{2} - 4ac = (-38)^{2} - 4 \cdot 21 \cdot 5 = 1444 - 420 = 1024 \] Now, substitute the values into the quadratic formula: \[ x = \frac{-(-38) \pm \sqrt{1024}}{2 \cdot 21} = \frac{38 \pm 32}{42} \] This gives us two possible solutions: 1. \( x = \frac{38 + 32}{42} = \frac{70}{42} = \frac{5}{3} \) 2. \( x = \frac{38 - 32}{42} = \frac{6}{42} = \frac{1}{7} \) Thus, the solutions are: \[ x = \frac{5}{3} \quad \text{and} \quad x = \frac{1}{7} \]