Solve 1/2left(2x-5right)^2-2=0

$$\frac{ 1 }{ 2 } { \left(2x-5 \right) }^{ 2 } -2=0$$

Solve for x

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps by Finding Square Root

Add 2 to both sides of the equation.

$$\frac{1}{2}\left(2x-5\right)^{2}-2+2=2$$

Subtracting 2 from itself leaves 0.

$$\frac{1}{2}\left(2x-5\right)^{2}=2$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}\left(2x-5\right)^{2}}{\frac{1}{2}}=\frac{2}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$\left(2x-5\right)^{2}=\frac{2}{\frac{1}{2}}$$

Divide 2 by \frac{1}{2} by multiplying 2 by the reciprocal of \frac{1}{2}.

$$\left(2x-5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$2x-5=2 2x-5=-2$$

Add 5 to both sides of the equation.

$$2x-5-\left(-5\right)=2-\left(-5\right) 2x-5-\left(-5\right)=-2-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$2x=2-\left(-5\right) 2x=-2-\left(-5\right)$$

Subtract -5 from 2.

$$2x=7$$

Subtract -5 from -2.

$$2x=3$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{7}{2} \frac{2x}{2}=\frac{3}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{7}{2} x=\frac{3}{2}$$