Solve 2(-4-x)+x^2<left(x-1right)^2-6

$$2(-4-x)+ { x }^{ 2 } < { \left(x-1 \right) }^{ 2 } -6$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply 2 by -4-x.

$$-8-2x+x^{2}<\left(x-1\right)^{2}-6$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$-8-2x+x^{2}<x^{2}-2x+1-6$$

Subtract 6 from 1 to get -5.

$$-8-2x+x^{2}<x^{2}-2x-5$$

Subtract x^{2} from both sides.

$$-8-2x+x^{2}-x^{2}<-2x-5$$

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

$$-8-2x<-2x-5$$

Add 2x to both sides.

$$-8-2x+2x<-5$$

Combine -2x and 2x to get 0.

$$-8<-5$$

This is true for any x.

$$x\in \mathrm{R}$$