How do you multiply #(n^2)/(2)-(5n)/(6) -2 =0#?

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Answer 1 · 2015-05-09T12:15:06

#n^2/2# #- (5n)/6# #- 2 = 0#

Here the L.C.M = 6

#(n^2 xx 3)/(2 xx 3)# #- (5n)/6# #- (2xx6)/(1 xx6) = 0#

#(3n^2)/(6)# #- (5n)/6# #- (12)/(6) = 0#
#(3n^2 - 5n - 12)/(6) = 0#

#3n^2 - 5n - 12 =6 xx 0#

#3n^2 - 5n - 12 = 0#

factoring by grouping / splitting the middle term:
#3n^2 - 5n - 12 = 0#

(# 3 xx -12 = -36# , #-36 = -9 xx 4# and #-5# = # 4-9#)

so,
#3n^2 - 9n + 4n - 12 = 0#

taking out the common terms:
#3n( n -3) + 4(n -3) = 0#
on grouping:
#(3n+4) (n - 3) = 0#

here, #n# has two solutions :
# n = 3 and n = -4/3#