Solve (2k+3)^2=8

$$( 2 k + 3 ) ^ { 2 } = 8$$

Solve for k

$$k=\sqrt{2}-\frac{3}{2}\approx -0.085786438$$

$$k=-\sqrt{2}-\frac{3}{2}\approx -2.914213562$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$2k+3=2\sqrt{2} 2k+3=-2\sqrt{2}$$

Subtract 3 from both sides of the equation.

$$2k+3-3=2\sqrt{2}-3 2k+3-3=-2\sqrt{2}-3$$

Subtracting 3 from itself leaves 0.

$$2k=2\sqrt{2}-3 2k=-2\sqrt{2}-3$$

Subtract 3 from 2\sqrt{2}.

$$2k=2\sqrt{2}-3$$

Subtract 3 from -2\sqrt{2}.

$$2k=-2\sqrt{2}-3$$

Divide both sides by 2.

$$\frac{2k}{2}=\frac{2\sqrt{2}-3}{2} \frac{2k}{2}=\frac{-2\sqrt{2}-3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$k=\frac{2\sqrt{2}-3}{2} k=\frac{-2\sqrt{2}-3}{2}$$

Divide 2\sqrt{2}-3 by 2.

$$k=\sqrt{2}-\frac{3}{2}$$

Divide -2\sqrt{2}-3 by 2.

$$k=-\sqrt{2}-\frac{3}{2}$$

The equation is now solved.

$$k=\sqrt{2}-\frac{3}{2} k=-\sqrt{2}-\frac{3}{2}$$