Solve left(-2xright)^2-4(x+1)(x-2)<0

$${ \left(-2x \right) }^{ 2 } -4(x+1)(x-2) < 0$$

Solve for x

$$x<-2$$

Solution Steps

Expand \left(-2x\right)^{2}.

$$\left(-2\right)^{2}x^{2}-4\left(x+1\right)\left(x-2\right)<0$$

Calculate -2 to the power of 2 and get 4.

$$4x^{2}-4\left(x+1\right)\left(x-2\right)<0$$

Use the distributive property to multiply -4 by x+1.

$$4x^{2}+\left(-4x-4\right)\left(x-2\right)<0$$

Use the distributive property to multiply -4x-4 by x-2 and combine like terms.

$$4x^{2}-4x^{2}+4x+8<0$$

Combine 4x^{2} and -4x^{2} to get 0.

$$4x+8<0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$4x<-8$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x<\frac{-8}{4}$$

Divide -8 by 4 to get -2.

$$x<-2$$