Does anyone know how to do this?

2 [ x+ h ]² + 6 [x + h] - [ 2x² + 6x ]
------------------------------------------------
h

2x² + 4hx + 2h² + 6x + 6h - 2x² - 6x
---------------------------------------------
h

4hx + 2h² + 6h
-------------------
h

4x + 2h + 6

4x + 6 ______as h-->0

Plug f(x) into the difference quotient.
f(x+h) - f(x) / h
((2(x+h)^2 + 6(x+h) - (2x^2 + 6x))/h.
(2x^2 + 4hx + 2h^2 + 6x + 6h - 2x^2 - 6x)/h
(4hx + 2h^2 + 6h)/h
4x + 2h + 6

You have:

f(x) = 2x² + 6x

And want to solve for:

[f(x + h) - f(x)] / h

So, substitute (x + h) in for x for the first function, then leave the second function as-is, and simplify:

[(2(x + h)² + 6(x + h)) - (2x² + 6x)] / h

Simplify:

[(2(x² + 2xh + h²) + 6x + 6h - (2x² + 6x)] / h
(2x² + 4xh + 2h² + 6x + 6h - 2x² - 6x) / h

Terms now start cancelling out;

(4xh + 2h² + 6h) / h

Factor out an h:

h(4x + 2h + 6) / h

And cancel :

4x + 2h + 6