Part 2 of 2 Solve the quadratic equation by completing the square. \[ x^{2}-6 x=6 \] Choose the appropriate process for rearranging the equation, if needed, before completing the square to obtain an equation of the form \( (x+k)^{2}=d \). (Type an integer or a simplified fraction.) A. \( x^{2}-6 x=6 \Rightarrow x^{2}-6 x+\square=6 \) B. \( x^{2}-6 x=6 \Rightarrow x^{2}-6 x+9=6+9 \) C. \( x^{2}-6 x=6 \Rightarrow \frac{1}{6} x^{2}-x+\square=1 \) D. \( x^{2}-6 x=6 \Rightarrow \frac{1}{6} x^{2}-x+\square=1+\square \) The solution set is (Type an exact answer, using radicals as needed. Use a comma to separate answers as needed. )

Let's start with rearranging the equation to get it ready for completing the square. Your initial equation is \(x^2 - 6x = 6\). To successfully complete the square, we want to shift it into a format like \( (x+k)^2 = d \). Among the given options, the correct one is B: \[ x^{2}-6 x=6 \Rightarrow x^{2}-6 x+9=6+9. \] By adding \( 9 \) to both sides, you complete the square on the left side, turning it into: \[ (x-3)^2 = 15. \] Now, solving for \( x \) gives \( x - 3 = \pm\sqrt{15} \), or: \[ x = 3 \pm \sqrt{15}. \] So, the solution set is: \[ 3 + \sqrt{15}, 3 - \sqrt{15}. \] For a bit of fun trivia, did you know that completing the square has ancient roots? It was used by mathematicians like Al-Khwarizmi in the 9th century, who laid down the groundwork for algebra! It's amazing how these age-old techniques still help us solve problems today. If you want to dig deeper into the elegance of quadratic equations, consider exploring more about the history of mathematics. Books or articles focused on ancient mathematical methods often show how these techniques evolved. It's fascinating to see how they still influence modern problem-solving!