college algebra

3y^2+8=0
(3y+2)+(3y-2)=8+0
3y=-8
y=3/8
this chapter is hard for me.

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1. No, it is wrong.

3y^2=-8
y^2=-8/3

Now you can't to anymore in the real number system, as there is no such thing as a square root of a negative number. But if you define i as sqrt (-1), then
y=+- sqrt (8/3) * i

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bobpursley

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