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# Quadratic equations help

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1. The question I have been asked to solve is:
y^2 - 5y = 3y^2 - 6y - 10

I have rearranged it to get -2y^2 + y + 10 = 0 but I don't understand how to factorise this as nothing multiplies to get 10 and adds to get 1
2. (Original post by bulbs)
The question I have been asked to solve is:
y^2 - 5y = 3y^2 - 6y - 10

I have rearranged it to get -2y^2 + y + 10 = 0 but I don't understand how to factorise this as nothing multiplies to get 10 and adds to get 1
factors of -20 that add to make 1
3. (Original post by bulbs)
The question I have been asked to solve is:
y^2 - 5y = 3y^2 - 6y - 10

I have rearranged it to get -2y^2 + y + 10 = 0 but I don't understand how to factorise this as nothing multiplies to get 10 and adds to get 1
Try to get the coefficient to be positive, by multiplying everything by -1. It should be easier from there.
4. (Original post by ValerieKR)
factors of -20 that add to make 1
Thanks.
So -2y^2 - 4y - 5y + 10 = 0
-2y(y+2) + 5(y+2)
(-2y+5)(y+2) = 0
y = 5/2 or y = -2
5. (Original post by bulbs)
Thanks.
So -2y^2 - 4y - 5y + 10 = 0
-2y(y+2) + 5(y+2)
(-2y+5)(y+2) = 0
y = 5/2 or y = -2
You've mixed up some signs in your type up but it's right

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