Solve left(2x+5right)^2=20

$${ \left(2x+5 \right) }^{ 2 } = 20$$

Solve for x

$$x=\sqrt{5}-\frac{5}{2}\approx -0.263932023$$

$$x=-\sqrt{5}-\frac{5}{2}\approx -4.736067977$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$2x+5=2\sqrt{5} 2x+5=-2\sqrt{5}$$

Subtract 5 from both sides of the equation.

$$2x+5-5=2\sqrt{5}-5 2x+5-5=-2\sqrt{5}-5$$

Subtracting 5 from itself leaves 0.

$$2x=2\sqrt{5}-5 2x=-2\sqrt{5}-5$$

Subtract 5 from 2\sqrt{5}.

$$2x=2\sqrt{5}-5$$

Subtract 5 from -2\sqrt{5}.

$$2x=-2\sqrt{5}-5$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{2\sqrt{5}-5}{2} \frac{2x}{2}=\frac{-2\sqrt{5}-5}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{2\sqrt{5}-5}{2} x=\frac{-2\sqrt{5}-5}{2}$$

Divide 2\sqrt{5}-5 by 2.

$$x=\sqrt{5}-\frac{5}{2}$$

Divide -2\sqrt{5}-5 by 2.

$$x=-\sqrt{5}-\frac{5}{2}$$

The equation is now solved.

$$x=\sqrt{5}-\frac{5}{2} x=-\sqrt{5}-\frac{5}{2}$$