what do you mean by "solve" -- describe its graph? find its zeroes?
Vertical asymptote at y = 8
zeroes at y = -3±√5
it can be written as y+14 - 116/(y-8), so
If z = (y^2+6y+4)/(y-8) then there is a slope asymptote at z = y+14
local max/min at y = 8±2√29
that enough for you?
How to solve: (y^2+6y+4)/(y-8)
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