document.write( "Question 257877: 2y(cubed)+y(squared)=3y
\n" ); document.write( "i subtracted 3y and got 2y(cubed)+y(squared)-3y=0
\n" ); document.write( "then i divided by y and got 2y(squared)+y-3=0
\n" ); document.write( "and then i got stuck
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Algebra.Com's Answer #189760 by richwmiller(17219)\"\" \"About 
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cubed is written ^3
\n" ); document.write( "squared is written ^2
\n" ); document.write( "2y^3+y^2=3y
\n" ); document.write( "y*(2y^2+y-3)=0
\n" ); document.write( "y*(y-1)*(2y+3) = 0
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