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.