Question 384896
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Hi,         
solving
3y^4=28y^2-9
3y^4 - 28y^2+ 9 = 0
Factoring
(3y^2 -1)(y^2 - 9)= 0 Note: inner and outer products = -28y^2
(3y^2 -1)= 0
  y^2 = 1/3
  y = ± {{{sqrt(1/3)}}} Or ± {{{sqrt(3)}}}/3
(y^2 - 9)= 0 
  y^2 = 9
  y =  ± 3