Solve left(2x-3right)^2=6

$${ \left(2x-3 \right) }^{ 2 } =6$$

Solve for x

$$x = \frac{\sqrt{6} + 3}{2} \approx 2.724744871$$

$$x=\frac{3-\sqrt{6}}{2}\approx 0.275255129$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$2x-3=\sqrt{6} 2x-3=-\sqrt{6}$$

Add 3 to both sides of the equation.

$$2x-3-\left(-3\right)=\sqrt{6}-\left(-3\right) 2x-3-\left(-3\right)=-\sqrt{6}-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$2x=\sqrt{6}-\left(-3\right) 2x=-\sqrt{6}-\left(-3\right)$$

Subtract -3 from \sqrt{6}.

$$2x=\sqrt{6}+3$$

Subtract -3 from -\sqrt{6}.

$$2x=3-\sqrt{6}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{\sqrt{6}+3}{2} \frac{2x}{2}=\frac{3-\sqrt{6}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{\sqrt{6}+3}{2} x=\frac{3-\sqrt{6}}{2}$$