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We have:

\(\frac{x}{x-5}=\frac{4}{x-4}+\frac{12}{x^2-9x+20}\)

The best way to approach this is to get rid of the fractions.

We notice that x²-9x+20 = (x-4)(x-5)

Now we multiply both sides by x²-9x+20

\((x^2-9x+20)(\frac{x}{x-5})=(x^2-9x+20)(\frac{4}{x-4}+\frac{12}{x^2-9x+20})\)

This equals:

\(x(x-4)=4(x-5)+12\)

\(x^2-4x=4x-20+12\)

\(x^2-8x+8=0\)

We notice this is not factorable, so we use the quadratic formula:  (+- means plus minus)

\({x}=\frac{8+-\sqrt{32}}{2}\)

\({x}_{1,2}=4+2\sqrt{2}, 4-2\sqrt{2}\)