The function $\displaystyle f$ is defined as follows :

$\displaystyle f(p) = 6p^2 - 4p$

Basically, you are trying to simplify this :

$\displaystyle \frac{f(5+h)-f(5)}{h}$ , $\displaystyle h \neq 0$

It is equivalent to (by substituting the function into your expression) :

$\displaystyle \frac{[6(5 + h)^2 - 4(5 + h)] - [6(5)^2 - 4(5)]}{h}$ , $\displaystyle h \neq 0$

Expand this :

$\displaystyle \frac{[6(5^2 + 10h + h^2) - 20 - 4h] - (150 - 20)}{h}$ , $\displaystyle h \neq 0$

Further expansion required ...

$\displaystyle \frac{150 + 60h + 6h^2 - 20 - 4h - 130}{h}$ , $\displaystyle h \neq 0$

Simplify :

> get rid of the constants (150, 130, 20, ...), they cancel out (who would have guessed it ?).

> simplify the linear coefficients (60h - 4h) !

> reorganize the equation, looks better ...

$\displaystyle \frac{6h^2 + 56h}{h}$ , $\displaystyle h \neq 0$

Factorize with $\displaystyle h$ :

$\displaystyle \frac{h(6h + 56)}{h}$ , $\displaystyle h \neq 0$

Finally, cancel out the $\displaystyle h$, to find :

$\displaystyle 6h + 56$

And you are done

EDIT : ah Defunkt you win, I was busy checking my reply :/