Solve -x^2leq6*-40

$$- { x }^{ 2 } \leq 6 \times -40$$

Solve for x

$$x\in (-\infty,-4\sqrt{15}]\cup [4\sqrt{15},\infty)$$

Solution Steps

Multiply 6 and -40 to get -240.

$$-x^{2}\leq -240$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x^{2}\geq \frac{-240}{-1}$$

Fraction \frac{-240}{-1} can be simplified to 240 by removing the negative sign from both the numerator and the denominator.

$$x^{2}\geq 240$$

Calculate the square root of 240 and get 4\sqrt{15}. Rewrite 240 as \left(4\sqrt{15}\right)^{2}.

$$x^{2}\geq \left(4\sqrt{15}\right)^{2}$$

Inequality holds for |x|\geq 4\sqrt{15}.

$$|x|\geq 4\sqrt{15}$$

Rewrite |x|\geq 4\sqrt{15} as x\leq -4\sqrt{15}\text{; }x\geq 4\sqrt{15}.

$$x\leq -4\sqrt{15}\text{; }x\geq 4\sqrt{15}$$