document.write( "Question 392420: Solve for y, where y is a real number.\r
\n" ); document.write( "\n" ); document.write( "\"+sqrt+%28y+%2B+20+=+y%29+\"\r
\n" ); document.write( "\n" ); document.write( "If there is more than one solution, separate them with commas
\n" ); document.write( "

Algebra.Com's Answer #278519 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
\"sqrt%28y%2B20%29+=+y\"
\n" ); document.write( "Square both sides:
\n" ); document.write( "\"y+%2B+20+=+y%5E2\",
\n" ); document.write( "==> \"0+=+y%5E2+-+y+-+20\"
\n" ); document.write( "==> (y-5)(y+4) = 0 ==> y = 5, -4.
\n" ); document.write( "Y = 5 satisfies the original equation, while y = -4 does not.
\n" ); document.write( "Therefore the final answer is only y = 5.
\n" ); document.write( "
\n" );