<PRE><B>Question 256849</B>
How would you write:
(48/n + 3)*(n-2)-48
as a standard form of a quadratic equation?

standard form would be an^2+bn+c=0 which means we need to figure out what a and b and c are

(48/n + 3)*(n-2)-48=0 use foil (first outer inner last)
48/n * n + 48/n * (-2) + 3*n + 3*(-2) - 48 = 0
48 - 96/n + 3*n - 6 - 48 = 0
   - 96/n + 3*n - 54 = 0
   - 32/n + n - 18 = 0 (divided by 3)
   -32 + n^2 - 18n = 0 (multiplied by n)
        n^2-18n-32=0</PRE>