Solve (z+1)^2-(z-1)^2=8

$$( z + 1 ) ^ { 2 } - ( z - 1 ) ^ { 2 } = 8$$

Solve for z

$$z=2$$

Steps for Solving Linear Equation

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

$$z^{2}+2z+1-\left(z-1\right)^{2}=8$$

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

$$z^{2}+2z+1-\left(z^{2}-2z+1\right)=8$$

To find the opposite of z^{2}-2z+1, find the opposite of each term.

$$z^{2}+2z+1-z^{2}+2z-1=8$$

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

$$2z+1+2z-1=8$$

Combine 2z and 2z to get 4z.

$$4z+1-1=8$$

Subtract 1 from 1 to get 0.

$$4z=8$$

Divide both sides by 4.

$$z=\frac{8}{4}$$

Divide 8 by 4 to get 2.

$$z=2$$