Solve 2(2-z)=xsqrt{left(2-zright)^2+4}

$$2(2-z)=x \sqrt{ { \left(2-z \right) }^{ 2 } +4 }$$

Solve for x

$$x=\frac{2\left(2-z\right)}{\sqrt{z^{2}-4z+8}}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by 2-z.

$$4-2z=x\sqrt{\left(2-z\right)^{2}+4}$$

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

$$4-2z=x\sqrt{4-4z+z^{2}+4}$$

Add 4 and 4 to get 8.

$$4-2z=x\sqrt{8-4z+z^{2}}$$

Swap sides so that all variable terms are on the left hand side.

$$x\sqrt{8-4z+z^{2}}=4-2z$$

The equation is in standard form.

$$\sqrt{z^{2}-4z+8}x=4-2z$$

Divide both sides by \sqrt{8-4z+z^{2}}.

$$\frac{\sqrt{z^{2}-4z+8}x}{\sqrt{z^{2}-4z+8}}=\frac{4-2z}{\sqrt{z^{2}-4z+8}}$$

Dividing by \sqrt{8-4z+z^{2}} undoes the multiplication by \sqrt{8-4z+z^{2}}.

$$x=\frac{4-2z}{\sqrt{z^{2}-4z+8}}$$

Divide 4-2z by \sqrt{8-4z+z^{2}}.

$$x=\frac{2\left(2-z\right)}{\sqrt{z^{2}-4z+8}}$$

Solve for z

$$z=-2\sqrt{-\frac{1}{x^{2}-4}}x+2$$

$$x>-2\text{ and }x<2$$