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y'' + ky' + (g/L)y = 0

I know how to find the solution for a simple pendulum without friction:

y'' = -(g/L)y ... which leads to ... y = Acos((g/L)x)

So far I have: y'' + ky' + (g/L)y = 0, substituting m = y' and w^2 = g/L,

m^2 + km + w^2 = 0 so m = [-k ± sqrt(k^2 - 4(1)(w^2))]/2 ... I'm stuck here because

I can't remember how to reduce this so I can find the homogeneous equation for y.