Solve for x
x\in (-\infty,-\sqrt{10}]\cup [\sqrt{10},\infty)
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-10+x^{2}\geq 0
Multiply the inequality by -1 to make the coefficient of the highest power in 10-x^{2} positive. Since -1 is negative, the inequality direction is changed.
x^{2}\geq 10
Add 10 to both sides.
x^{2}\geq \left(\sqrt{10}\right)^{2}
Calculate the square root of 10 and get \sqrt{10}. Rewrite 10 as \left(\sqrt{10}\right)^{2}.
|x|\geq \sqrt{10}
Inequality holds for |x|\geq \sqrt{10}.
x\leq -\sqrt{10}\text{; }x\geq \sqrt{10}
Rewrite |x|\geq \sqrt{10} as x\leq -\sqrt{10}\text{; }x\geq \sqrt{10}.
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