We'll decompose the fraction into elementary irreducible fractions:

(5x-4)/(x-2)(x+3) = A/(x-2) + B/(x+3)

We'll calculate LCD of the 2 fractions from the right side.

The LCD is the same with the denominator from the left side.

LCD = (x-2)(x+3)

We'll multiply both sides by (x-2)(x+3) and the expression will become:

(5x-4) = A(x+3) + B(x-2)

We'll remove the brackets:

5x - 4 = Ax + 3A + Bx - 2B

We'll combine like terms form the right side:

5x - 4 = x(A+B) + (3A-2B)

Comparing we'll get:

5 = A+B

-4 = 3A-2B

We'll use the symmetric property:

A+B = 5 (1)

3A-2B = -4 (2)

We'll multiply (1) by 2:

2A+2B = 10 (3)

We'll add (3) to (2):

2A+2B+3A - 2B = 10-4

We'll eliminate like terms:

5A = 6

We'll divide by 5:

A = 6/5

We'll substitute A in (1):

A+B = 5

6/5 + B = 5

We'll subtract 6/5 both sides:

B = 5 - 6/5 => B = (25-6)/5 => B = 19/5

**The given fraction written as partial fractions is: (5x-4)/(x-2)(x+3) = 6/5(x-2) + 19/5(x+3), where ****6/5(x-2)**** and ****19/5(x+3)**** are partial fractions.**