Solve -2x^2<-5

$$-2 { x }^{ 2 } < -5$$

Solve for x

$$x\in \left(-\infty,-\frac{\sqrt{10}}{2}\right)\cup \left(\frac{\sqrt{10}}{2},\infty\right)$$

Solution Steps

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

$$x^{2}>\frac{-5}{-2}$$

Fraction \frac{-5}{-2} can be simplified to \frac{5}{2} by removing the negative sign from both the numerator and the denominator.

$$x^{2}>\frac{5}{2}$$

Calculate the square root of \frac{5}{2} and get \frac{\sqrt{10}}{2}. Rewrite \frac{5}{2} as \left(\frac{\sqrt{10}}{2}\right)^{2}.

$$x^{2}>\left(\frac{\sqrt{10}}{2}\right)^{2}$$

Inequality holds for |x|>\frac{\sqrt{10}}{2}.

$$|x|>\frac{\sqrt{10}}{2}$$

Rewrite |x|>\frac{\sqrt{10}}{2} as x<-\frac{\sqrt{10}}{2}\text{; }x>\frac{\sqrt{10}}{2}.

$$x<-\frac{\sqrt{10}}{2}\text{; }x>\frac{\sqrt{10}}{2}$$