How do you solve #12n ^ { 2} + 2n = 4#?

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Answer 1 · 2016-12-02T17:49:48

#n = 1/2 or n = -2/3#

When solving a quadratic you generally make it equal to 0.

#12n^2 +2n -4 = 0" "larr div 2#

#6n^2 +n -2 = 0" "larr# find its factors

Find factors of 6 and 2 which subtract to make 1.
The brackets will have different signs.
The larger term in #n# must be positive.

#(3n +2)(2n-1) =0#

Set each factor equal to 0.

#3n +2 = 0#
#3n = -2#
#n = -2/3#

#2n-1 =0#
#2n =1#
#n = 1/2#