A student solved the equation x^2-6x+6=0 by completing the square. Move numbers to the lines to complete the student's calculation. x^2-6x+6=0 (x^2-6x+underline ( ))+underline ( )+6=0 (x-3)^2=underline ( ) x-3=pm underline ( ) x=underline ( )+underline ( ) and x x=underline ( )+underline ( ) -9 -6 -3

To complete the square for the equation \(x^{2}-6 x+6=0\), we need to find a number to fill in the square that makes the left side a perfect square trinomial.The equation is \(x^{2}-6 x+6=0\).Step 1: Take half of the coefficient of \(x\), which is \(-6\), and square it. This gives us \((-3)^2 = 9\).Step 2: Add and subtract this number inside the parentheses to complete the square.\[\begin{array}{l}x^{2}-6 x+9-9+6=0 \\\left(x^{2}-6 x+9\right)-3=0 \\(x-3)^{2}-3=0 \\\end{array}\]Step 3: Solve for \(x\) by taking the square root of both sides.\[\begin{array}{l}(x-3)^{2}=3 \\x-3= \pm \sqrt{3} \\\end{array}\]Step 4: Solve for the two possible values of \(x\).\[\begin{array}{l}x=3+\sqrt{3} \\x=3-\sqrt{3} \\\end{array}\]So the completed calculation is:\[\begin{array}{l}x^{2}-6 x+6=0 \\\left(x^{2}-6 x+9\right)-3=0 \\(x-3)^{2}=3 \\x-3= \pm \sqrt{3} \\x=3+\sqrt{3} \\x=3-\sqrt{3} \\\end{array}\]Therefore, the numbers that fill in the blanks are:\[\begin{array}{l}\boxed{9} \text{ for the square} \\\boxed{\sqrt{3}} \text{ for the square root} \\\boxed{3} \text{ for the constant term} \\\end{array}\]And the solutions are:\[x=3+\sqrt{3} \quad \text{and} \quad x=3-\sqrt{3}\]