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

To solve the equation \( x^{2}=(x-1)^{2}+(x-2)^{2} \), start by expanding the right-hand side: \[ (x-1)^{2} = x^{2} - 2x + 1 \quad \text{and} \quad (x-2)^{2} = x^{2} - 4x + 4. \] So, \[ (x-1)^{2} + (x-2)^{2} = (x^{2} - 2x + 1) + (x^{2} - 4x + 4) = 2x^{2} - 6x + 5. \] Now, set the left-hand side equal to the expanded right-hand side: \[ x^{2} = 2x^{2} - 6x + 5. \] Rearranging gives: \[ 0 = 2x^{2} - x^{2} - 6x + 5 \implies 0 = x^{2} - 6x + 5. \] Now, apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b = -6 \), and \( c = 5 \): \[ x = \frac{-(-6) \pm \sqrt{(-6)^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} = \frac{6 \pm 4}{2}. \] This gives us two possible solutions: \[ x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{2}{2} = 1. \] So, the solutions are \( x = 5 \) and \( x = 1 \).