Find the infimum and supremum of {x ∈ R : x2+2x=3}
"\\{x \u2208 R : x^2+2x=3\\}=\\{x \u2208 R : (x+1)^2=4\\}=\\{1, -3\\}"
Therefore, "\\inf \\{x \u2208 R : x^2+2x=3\\}=\\inf\\{1, -3\\}=-3",
"\\sup \\{x \u2208 R : x^2+2x=3\\}=\\sup\\{1, -3\\}=1"
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