solve this equation for using the quadratic formula \( x^{2}=(x-1)^{2}+(x-2)^{2} \), and \( x ? \)

To solve the equation \( x^{2} = (x-1)^{2} + (x-2)^{2} \), let's first expand the right side: 1. Expanding \( (x-1)^{2} \) gives \( x^{2} - 2x + 1 \). 2. Expanding \( (x-2)^{2} \) gives \( x^{2} - 4x + 4 \). Now combine these: \[ (x-1)^{2} + (x-2)^{2} = (x^{2} - 2x + 1) + (x^{2} - 4x + 4) = 2x^{2} - 6x + 5 \] Now, substitute this back into the original equation: \[ x^{2} = 2x^{2} - 6x + 5 \] Rearranging gives: \[ 0 = 2x^{2} - 6x + 5 - x^{2} \] \[ 0 = x^{2} - 6x + 5 \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = 5 \). Calculating the discriminant: \[ b^{2} - 4ac = (-6)^{2} - 4(1)(5) = 36 - 20 = 16 \] Now substituting into the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{16}}{2(1)} = \frac{6 \pm 4}{2} \] This gives us two solutions: 1. \( x = \frac{10}{2} = 5 \) 2. \( x = \frac{2}{2} = 1 \) Thus, the solutions to the equation are \( x = 5 \) and \( x = 1 \).