Solve for x
x\in \left(-\infty,-\sqrt{6}\right)\cup \left(\sqrt{6},\infty\right)
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2x^{2}-12>0
Subtract 17 from 5 to get -12.
x^{2}>6
Add 6 to both sides.
x^{2}>\left(\sqrt{6}\right)^{2}
Calculate the square root of 6 and get \sqrt{6}. Rewrite 6 as \left(\sqrt{6}\right)^{2}.
|x|>\sqrt{6}
Inequality holds for |x|>\sqrt{6}.
x<-\sqrt{6}\text{; }x>\sqrt{6}
Rewrite |x|>\sqrt{6} as x<-\sqrt{6}\text{; }x>\sqrt{6}.
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