#2x^2-10x=-12#
to complete the square of an expression:
#ax^2 + bx +c#
#a=1#
#c=(b/2)^2#
So in your equation we need to divide both sides by 2 to get rid of the coefficient of #x^2#
#(2x^2-10x)/2=(-12)/2#
#x^2-5x=-6#
#a=1#
#b=-5#
now put your #c# in, remember you have to add it to both sides of the equation:
#x^2-5x + c=-6 +c#
#c=(b/2)^2= ((-5)/2)^2 =25/4#
#x^2-5x + 25/4=-6 +25/4#
finally the square goes like this:
#(x + b/2)^2=-6 +25/4#
#(x + (-5)/2)^2 = -6 +25/4#
#(x -5/2)^2 = 1/4#
#sqrt((x -5)/2^2) = +-sqrt(1/4)#
#x -5/2 = +-1/2#
#x = 5/2+-1/2#
#x=2 and x = 3#
Just as a note this would be much easier to just factor the normal way; when the #b# is not evenly divisible by 2 as in this case #b=-5# it is much easier to use the quadratic formula if you can't just factor normally or by grouping.
