#3-sqrt(3n²+n-3)=n+1# how do you solve this equation ?

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Answer 1 · 2017-05-14T10:37:21

#n = (-7/2,1)#

#3-sqrt(3n²+n-3)=n+1rArr sqrt(3n²+n-3)=2-n#

now squaring both sides

#3n²+n-3=4-4n+n^2#

simplifying

#2 n^2 + 5 n - 7=0#

and solving for #n#

#n = (-7/2,1)# both are feasible solutions because for them #3n²+n-3 > 0#