Let z=9/(y(y-2))-y²+27 and sketch the graph, where y is the horizontal axis. Note that z is not defined for y=0, 2.
From the graph we can see that there is a zero (y value) at about -5, 0, 2 and 5.
We can use calculus and Newton’s Method to find accurate values for solutions for y. But two of the zeroes are close to the prohibited values 0 and 2. Therefore we need an equation where this would not be a problem. We can redefine z=y⁴-2y³-27y²+54y-9
Newton’s Method:
y=y-z/z', where z'=4y³-6y²-54y+54.
y=y-(y⁴-2y³-27y²+54y-9)/(4y³-6y²-54y+54), iterative formula which uses a value of y on the right-hand side close to a zero to find a more accurate value.
Initial y=-5: final y=-5.219087298
Initial y=0: final y=0.1837590655
Initial y=2: final y=1.788583622
Initial y=5: final y=5.24674461