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one, and they are teh same number.

well, two, but they are the the same number is what I mean.

We can let \(y = 0\) and then solve \(2x^2 + 4x + 2 = 0\).

sorry, its -1

2x^2+2x+1+0
or, x^2_2x+1=0
or, (x+1)^2=0
or, X+1=0
or, x= -1

sorry, 2x^2+4x+2=0 *

thanks guys !