document.write( "Question 31435: solve for y:6y^2+y=12 \n" ); document.write( "

Algebra.Com's Answer #18157 by ikdeep(226) $\"\"$ $\"About$

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6y^2 + y = 12\r
\n" ); document.write( "\n" ); document.write( "6y^2 + y - 12 = 0 \r
\n" ); document.write( "\n" ); document.write( "Here you must know how we make factors i.e. splitting of the middle term...if you might not knew then contact me again and I will explain to you)... \r
\n" ); document.write( "\n" ); document.write( "6y^2 + (9-8)y - 12 = 0 \r
\n" ); document.write( "\n" ); document.write( "opening brackets would give ,,, \r
\n" ); document.write( "\n" ); document.write( "6y^2 + 9y - 8y - 12 = 0 \r
\n" ); document.write( "\n" ); document.write( "taking common from first two i.e. 6y^2 + 9y and last two i.e. - 8y - 12 variables...we get \r
\n" ); document.write( "\n" ); document.write( "3y(2y+3) - 4(2y+3) = 0\r
\n" ); document.write( "\n" ); document.write( "(3y-4)(2y+3)= 0\r
\n" ); document.write( "\n" ); document.write( "so either y = 4/3\r
\n" ); document.write( "\n" ); document.write( "or y = -3/2\r
\n" ); document.write( "\n" ); document.write( "hope this will help you
\n" ); document.write( "Please feel free to revert back for any further queries. \r
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