Solve (2x-1)^2-9=1

$$( 2 x - 1 ) ^ { 2 } - 9 = 1$$

Solve for x

$$x = \frac{\sqrt{10} + 1}{2} \approx 2.08113883$$

$$x=\frac{1-\sqrt{10}}{2}\approx -1.08113883$$

Steps by Finding Square Root

Add 9 to both sides of the equation.

$$\left(2x-1\right)^{2}-9+9=1+9$$

Subtracting 9 from itself leaves 0.

$$\left(2x-1\right)^{2}=1+9$$

Add 1 to 9.

$$\left(2x-1\right)^{2}=10$$

Take the square root of both sides of the equation.

$$2x-1=\sqrt{10} 2x-1=-\sqrt{10}$$

Add 1 to both sides of the equation.

$$2x-1-\left(-1\right)=\sqrt{10}-\left(-1\right) 2x-1-\left(-1\right)=-\sqrt{10}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$2x=\sqrt{10}-\left(-1\right) 2x=-\sqrt{10}-\left(-1\right)$$

Subtract -1 from \sqrt{10}.

$$2x=\sqrt{10}+1$$

Subtract -1 from -\sqrt{10}.

$$2x=1-\sqrt{10}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\sqrt{10}+1}{2} \frac{2x}{2}=\frac{1-\sqrt{10}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\sqrt{10}+1}{2} x=\frac{1-\sqrt{10}}{2}$$