Question 96676
completing the square means turning the quadratic into a "perfect square"


first, (using addition or subtraction) move the constant term to the other side of the equation


x^2-6x=3


next, (using multiplication or division) set the x^2 coefficient equal to 1 ... already done here


next, square 1/2 of the x term coefficient and add it to both sides ... -6/2=-3 ... (-3)^2=9


x^2-6x+9=12 ... (x-3)^2=12


next, take the square root of both sides ... remembering that square roots are + and -


x-3=sqrt(12) ... x-3=2(sqrt(3)) ... so x=3+2(sqrt(3)) and x=3-2(sqrt(3))