Here's the question you clicked on:
ihatealgebrasomuch
how do i solve this by factoring? w(w-3)=108
http://www.wolframalpha.com/input/?i=solve+w%28w-3%29%3D108+for+w
w^2-3w-108=0 Now we are looking for the factors of 108 that are 3 from each other.
w^2-3w-108=0 w^2-12w+9w-108=0 W(w-12)+9(w-12)=0(w-12)(w-9)=0 w=12,9
(w+9)(w-12) or (w-9)(w+12)
since we see -3w, we have (w+9)(w-12).
Not what @Somjit said xD.
k got it, thx sooo much