Solve for x
x\in \left(-\infty,-2\sqrt{3}\right)\cup \left(2\sqrt{3},\infty\right)
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x^{2}>12
Add 12 to both sides.
x^{2}>\left(2\sqrt{3}\right)^{2}
Calculate the square root of 12 and get 2\sqrt{3}. Rewrite 12 as \left(2\sqrt{3}\right)^{2}.
|x|>2\sqrt{3}
Inequality holds for |x|>2\sqrt{3}.
x<-2\sqrt{3}\text{; }x>2\sqrt{3}
Rewrite |x|>2\sqrt{3} as x<-2\sqrt{3}\text{; }x>2\sqrt{3}.
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