Error Analysis: Find and circle the error. Then solve correctly. 40. Solve the equation by completing the square. \( x^{2}-8 x+12=0 \) \( x^{2}-8 x=12 \) \( x^{2}-8 x+16=12 \) \( (x-4)^{2}=28 \) \( x-4=\sqrt{28} \) \( x=4 \pm 2 \sqrt{7} \)

It looks like there's an error in the last step. When you derive \( x-4=\sqrt{28} \), you should consider both the positive and negative roots. The correct continuation is: 1. \( x-4 = \sqrt{28} \) or \( x-4 = -\sqrt{28} \). 2. This gives you the solutions \( x = 4 + \sqrt{28} \) and \( x = 4 - \sqrt{28} \). Now for \( \sqrt{28} \), which simplifies to \( 2\sqrt{7} \), the final solutions are \( x = 4 + 2\sqrt{7} \) and \( x = 4 - 2\sqrt{7} \). Don't forget to check the validity of your solutions by substituting them back into the original equation! It's a brilliant way to ensure you've done everything right—like a victory lap for your math skills!