1.)rewrite in order to solve
3/2x-6 + 5/x-3 = 1/6
2.)rewrite in order to complete the square. (i hate these)
y^2 +3y-1=0
in future, please use grouping symbols for the sake of clarity ...
3/(2x-6) + 5/(x-3) = 1/6
3/[2(x-3)] + 5/(x-3) = 1/(2*3) note that x cannot equal 3.
get a common denominator ...
3*3/[3*2(x-3)] + 5*2*3/[2*3(x-3)] = 1(x-3)/[2*3(x-3)]
now that you have a common denominator, the numerators form the equation ...
9 + 30 = x - 3
now can you find x ?
y^2 + 3y - 1 = 0
y^2 + 3y = 1
half the coefficient of the linear term (that's the 3), square, and add result to both sides ...
y^2 + 3y + (3/2)^2 = 1 + (3/2)^2
[y + (3/2)]^2 = 13/4
y + (3/2) = +/- sqrt(13/4) = +/- sqrt(13)/2
y = [3 +/- sqrt(13)]/2
get over your hatred of these ... they are not going to go away.