This is not an eqn. You have to simply factorize the expression.
2y^2+12y-54
= 2 ( y^2 + 6y - 27)
=> 2 [ y^2 + 9y - 3y - 27 ]
=> 2 [ y ( y + 9 ) - 3 ( y + 9 ) ]
=> 2. ( y + 9 ) ( y - 3 ) ....................... Answer
2007-12-23 02:29:00
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answer #1
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answered by Pramod Kumar 7
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first take out the two
2(y^2+6y-27)
and then factor
2(y-3)(y+9)
then check it to make sure
so multiply it out y^2+9y-3y-27
combine y's and y^2+6y-27
then multiply by the two
2y^2+12y-54
which is the correct answer!
so
2(y-3)(y+9) is equivalent to 2y^2+12y-54
and other peeps
it doesn't ask for what y equals just the equivalent expression :)
2007-12-23 12:24:15
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answer #2
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answered by the helper 2
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first of all, this is only an EXPRESSION not an EQUATION,
so you cannot solve for y
you can factor it...
~ 2y^2 + 12y - 54
~ (2)( y^2 + 6y - 27) ....... find factors of -27 that add to 6, such
as 9 and -3
~ (2)(y+9)(y-3)........ <- answer
2007-12-23 15:31:26
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answer #3
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answered by c 2
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2y^2+12y+54
Multiply 54 into 2 and prime factorise it.
The factors will be 2,2,3,3,3.
Now try to make a combination that will give you 12.
The combination will be 3x3x2-3x2
This means 18-6
18-6 is 12 therfore take this combination.
Now you will get, 2y^2+18y-6y-54
Take the common factor out,that is
2y(y+9)-6(y+9)
therefore y=-9 or y=6/2=3
therefore y=-9,3
2007-12-23 08:48:53
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answer #4
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answered by Anirudh N 1
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2y^2+12y-54
(2y-6)(y+9)
2y-6=0
y+9=0
y=3 or -9
2007-12-23 12:22:40
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answer #5
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answered by Dave aka Spider Monkey 7
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You can check out www.gobakersfield.net and click on GoEducation, you will find math homework links and the famous Algebrator. This is a calculator that will do your homework step-by-step (your personal tutor)... Hope this help, as it has helped me.
2007-12-23 21:37:23
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answer #6
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answered by John R 2
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(2)(y+9)(y-3)
2007-12-23 10:31:16
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answer #7
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answered by Bruce J 1
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2(y^2+6y-27)
2(y+9)(y-3)
y= - 9,3
2007-12-23 08:40:26
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answer #8
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answered by Anonymous
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well if you're 8th grade or younger, i would go to www.aaamath.com or www.aplusmath.com. If you're college or HS level, i woul go to www.algebrahelp.com or www.geometry.com.
2007-12-23 08:41:45
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answer #9
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answered by chicagobeardude 1
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