Solve 1/2(2x-5)^2-8=0

$$\frac { 1 } { 2 } ( 2 x - 5 ) ^ { 2 } - 8 = 0$$

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$x=\frac{1}{2}=0.5$$

Steps by Finding Square Root

Add 8 to both sides of the equation.

$$\frac{1}{2}\left(2x-5\right)^{2}-8+8=8$$

Subtracting 8 from itself leaves 0.

$$\frac{1}{2}\left(2x-5\right)^{2}=8$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}\left(2x-5\right)^{2}}{\frac{1}{2}}=\frac{8}{\frac{1}{2}}$$

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

$$\left(2x-5\right)^{2}=\frac{8}{\frac{1}{2}}$$

Divide 8 by \frac{1}{2} by multiplying 8 by the reciprocal of \frac{1}{2}.

$$\left(2x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$2x-5=4 2x-5=-4$$

Add 5 to both sides of the equation.

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

Subtracting -5 from itself leaves 0.

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

Subtract -5 from 4.

$$2x=9$$

Subtract -5 from -4.

$$2x=1$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{9}{2} \frac{2x}{2}=\frac{1}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{9}{2} x=\frac{1}{2}$$