Sorry if it's a bit too late, you may have already solved the problem, but basically, you take the 3 in front of the first bracket, and times it by it to get (3x-9)(x+2) and then the same for the second part "-2(x+4)(x-5)" you'll get (-2x-8)(x-5). After that you just times the brackets out, so in the first bracket you times 3x by x and 2 of the second bracket you'll get 3x^2 +6x then take the -9 and do the same thing to get -9x-18

so now it looks like this :

3x^2+6x-9x-18+(-2x+4)(x-5) now do the same thing to the second pair of brackets and you get your answer. Cheers

PS: i'm not sure which section of Khan has lessons on this, but i'm pretty sure it's in there, somewhere in lower tier math.

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

foil

3*(x^2+2x-3x-6) -2*(x^2-5x+4x-20)=

simplify

3(x^2-x-6)-2(X^2-x+20)=

distribute (multiply by 3, then -2, not 2)

(3x^2-3x-18)+(-2x^2+2x-40)=

simplify

x^2-x+22

This is a classic example of foiling (multiplying a binomial by a binomial), then doing an operation and making sure the answer is simplified.  While it is possible to distribute the 3 and -2 through the terms, I prefer to simplify as much as possible as early as possible.

Alternate:

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

distribute (-2 NOT 2)

(3x-9)(x+2)+(-2x-8)(x-5)

foil

(3x^2-9x+6x-18)+(-2x^2-8x+10x+40)

simplify

x^2-x+22

X rise to the power 2 minus X minus 6=0