document.write( "Question 218354: solve the equation 2y^2-8y-5=0
\n" ); document.write( "you cant factor this out, i've tried to
\n" ); document.write( "im stuck on whats next, the home work says
\n" ); document.write( "1/2[2y^2-8y-5=0] then square both sides
\n" ); document.write( "why do you first use 1/2 with [2y^2-8y-5=0] ?
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Algebra.Com's Answer #164457 by Alan3354(69443)\"\" \"About 
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solve the equation 2y^2-8y-5=0
\n" ); document.write( "you cant factor this out, i've tried to
\n" ); document.write( "im stuck on whats next, the home work says
\n" ); document.write( "1/2[2y^2-8y-5=0] then square both sides
\n" ); document.write( "----------------
\n" ); document.write( "It might say multiply by 1/2, then complete the square of the eqn.
\n" ); document.write( "1/2[2y^2-8y-5=0]
\n" ); document.write( "--> y^2 - 4y = .25
\n" ); document.write( "Then add 4 (to both sides) to make the left side a square.
\n" ); document.write( "y^2 - 4y + 4 = 4.25
\n" ); document.write( "(y-2)^2 = 4.25
\n" ); document.write( "Now take sqrt of both sides
\n" ); document.write( "y-2 = sqrt(4.25)
\n" ); document.write( "y-2 = ~ 2.062
\n" ); document.write( "y = ~ 4.062
\n" ); document.write( "------------
\n" ); document.write( "y-2 = ~ -2.062
\n" ); document.write( "y = ~ 0.062
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