How do you solve for n in #(9n+4)+(3n^2 - 0.75) = 180#?

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Answer 1 · 2018-07-07T12:10:44

#n_1=sqrt(2202)/6-9/6# or #n_2=-9/6-sqrt(2202)/6#

Simplifying your equation

#3n^2+9n+4-3/4-180=0#

we have

#4-3/4-180=(16-3-360)/4=-707/4#

so we have to solve

#3n^2+9n-707/4=0#
dividing by #3# we get

#n^2+3n-707/12=0#

#n_(1,2)=-3/2\pm sqrt(367/6)#

this is

#n_(1,2)=-3/2pm sqrt(2202)/6#

so we get

#n_1=(sqrt(2202)-9)/6#

#n_2=-1/6(9+sqrt(2202))#