3y^2+5y=2

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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3y^2+5y-2=0 (3y-1)(y+2)
im confused how did you get -2 in the equation?
You subtract 2 from each side of the equation which then gives you a -2 on the left hand side of the equation and zero on the right hand side of the equation. Once you do this, you can then factor the equation.

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ooohh! okay thanks!
would the answer be y=1/3 and -2?
You're welcome. Hopefully, you understood what I did. Yes, y=1/3 and -2.
so for 4m^2 -100=0 would we do the same thing?
Basically, you would do the same thing. The factor in this problem would be: (2m-10)(2m+10) and m=5 and -5.
5x^2=40 this one were having alot of trouble on
5x^2 - 40 = 0 5(x^2-8) x=+- 2sqrt(2) ??
What you can do is to divide both sides by 5 and you get: x^2=8 x=square root of 8=+/- square root (4)(2)=+/-2(square root 2)
\[\sqrt{8} =+/-(4) (2)=+/-2\sqrt{2}\]
is that how you write it?
im so confused
im so confused
To be completely correct you should put a +/- in front of square root 8 and use the same symbol for the square roots of 4 and 2. Otherwise, you are correct. What is the confusion?

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