document.write( "Question 643590: Find the reals of the solution: 2=y+6y^2 \n" ); document.write( "

Algebra.Com's Answer #404556 by jsmallt9(3758) $\"\"$ $\"About$

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$\"2=y%2B6y%5E2\"$
\n" ); document.write( "To solve a quadratic equation, you want one side to be zero. Subtracting 2 from each side:
\n" ); document.write( " $\"0=y%2B6y%5E2-2\"$
\n" ); document.write( "Then we'll put the terms in \"proper\" order (highest exponent to lowest):
\n" ); document.write( " $\"0=6y%5E2%2By-2\"$
\n" ); document.write( "Now we factor:
\n" ); document.write( " $\"0+=+%283y%2B2%29%282y-1%29\"$
\n" ); document.write( "Next we use the Zero Product Property which tells us that for any product, like this one, to be zero one of its factors must be zero. So:
\n" ); document.write( " $\"3y%2B2+=+0\"$ or $\"2y-1=0\"$
\n" ); document.write( "Solving these...
\n" ); document.write( " $\"3y+=+-2\"$ or $\"2y=1\"$
\n" ); document.write( " $\"y+=+-2%2F3\"$ or $\"y=1%2F2\"$
\n" ); document.write( "Both of these are real numbers. \n" ); document.write( "

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