On how to solve binomials! show your work! (n-1)(2n-2)

Perhaps you want to multiply them out. You make use of the distributive property for that purpose.

That property tells you
   a(b + c) = ab + ac

For the purpose here, in the first step, you can consider either binomial to be "a". Suppose we choose a=(n-1), b=2n, c=-2. Then we have
   (n-1)(2n - 2) = (n-1)(2n) + (n-1)(-2)

Again, we have products that can make use of the distributive property. If you like, you can rearrange them so the monomial is in front.
   2n(n - 1) + (-2)(n - 1)
We can use the distributive property twice more to get
   2n(n) +2n(-1) +(-2)(n) +(-2)(-1)

Eliminating parentheses, this collection of terms becomes
   2n² -2n -2n +2
and we can combine the middle two "like" terms to get the product as
  2n² -4n +2


There is an acronym (FOIL) that is sometimes taught as a reminder how to multiply binomials. It applies very specifically to the case of multiplying two binomials. The distributive property applies in all cases.

FOIL tells you to sum the products of ...
  First terms (n -1)(2n -2) . . . . . . . 2n²
  Outer terms (n -1)(2n -2) . . . . .. -2n
  Inner terms (n -1)(2n -2) . . . . . . -2n
  Last terms (n -1)(2n -2) . . . . . . . 2
The sum is then 2n² -4n +2, as above.

AOS : x = -b/(2a)
x = -b/6
c is the constant.
c = 4