$\displaystyle \frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{x^2-4} = \frac{x+5}{x-2} = \frac{5}{x+2} + \frac{28}{(x-2)(x+2)}$

$\displaystyle \frac{x+5}{x-2} \cdot \frac{x+2}{x+2} = \frac{(x+5)(x+2)}{(x-2)(x+2)}$

$\displaystyle \frac{5}{x+2} \cdot \frac{x-2}{x-2} = \frac{5(x-2)}{(x-2)(x+2)}$

Combine all these now

$\displaystyle \frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x-2)(x+2)} + \frac{28}{x^2-4}$

$\displaystyle \frac{(x+5)(x+2)}{(x-2)(x+2)} = \frac{5(x-2)}{(x-2)(x+2)} + \frac{28}{(x-2)(x+2)}$

$\displaystyle (x-2)(x+2)$ will cancel

$\displaystyle (x+5)(x+2) = 5(x-2) + 28$

Expand using FOIL

$\displaystyle x^2+5x+2x+10 = 5x-10+28$

Simplify

$\displaystyle x^2+7x+10=5x+18$