Solve (6+2z)^2=6

$$( 6 + 2 z ) ^ { 2 } = 6$$

Solve for z

$$z=\frac{\sqrt{6}}{2}-3\approx -1.775255129$$

$$z=-\frac{\sqrt{6}}{2}-3\approx -4.224744871$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$2z+6=\sqrt{6} 2z+6=-\sqrt{6}$$

Subtract 6 from both sides of the equation.

$$2z+6-6=\sqrt{6}-6 2z+6-6=-\sqrt{6}-6$$

Subtracting 6 from itself leaves 0.

$$2z=\sqrt{6}-6 2z=-\sqrt{6}-6$$

Subtract 6 from \sqrt{6}.

$$2z=\sqrt{6}-6$$

Subtract 6 from -\sqrt{6}.

$$2z=-\sqrt{6}-6$$

Divide both sides by 2.

$$\frac{2z}{2}=\frac{\sqrt{6}-6}{2} \frac{2z}{2}=\frac{-\sqrt{6}-6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$z=\frac{\sqrt{6}-6}{2} z=\frac{-\sqrt{6}-6}{2}$$

Divide \sqrt{6}-6 by 2.

$$z=\frac{\sqrt{6}}{2}-3$$

Divide -\sqrt{6}-6 by 2.

$$z=-\frac{\sqrt{6}}{2}-3$$

The equation is now solved.

$$z=\frac{\sqrt{6}}{2}-3 z=-\frac{\sqrt{6}}{2}-3$$