Solve ∣x2−3∣<1|x^2 - 3| < 1∣x2−3∣<1.
(2,2)∪(−2,−2)(\sqrt{2}, 2) \cup (-2, -\sqrt{2})(2,2)∪(−2,−2)
(−2,2)(-\sqrt{2}, \sqrt{2})(−2,2)
(−2,2)(-2, 2)(−2,2)
(2,2)(\sqrt{2}, 2)(2,2)