Solve (x/2+3)(x/2-3)=(x/2-1)^2

$$( \frac { x } { 2 } + 3 ) ( \frac { x } { 2 } - 3 ) = ( \frac { x } { 2 } - 1 ) ^ { 2 }$$

Solve for x

$$x=10$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2\left(\frac{x}{2}+3\right)\left(\frac{x}{2}-3\right)=2\left(\frac{x}{2}-1\right)^{2}$$

Use the distributive property to multiply 2 by \frac{x}{2}+3.

$$\left(2\times \frac{x}{2}+6\right)\left(\frac{x}{2}-3\right)=2\left(\frac{x}{2}-1\right)^{2}$$

Express 2\times \frac{x}{2} as a single fraction.

$$\left(\frac{2x}{2}+6\right)\left(\frac{x}{2}-3\right)=2\left(\frac{x}{2}-1\right)^{2}$$

Cancel out 2 and 2.

$$\left(x+6\right)\left(\frac{x}{2}-3\right)=2\left(\frac{x}{2}-1\right)^{2}$$

Use the distributive property to multiply x+6 by \frac{x}{2}-3.

$$x\times \frac{x}{2}-3x+6\times \frac{x}{2}-18=2\left(\frac{x}{2}-1\right)^{2}$$

Express x\times \frac{x}{2} as a single fraction.

$$\frac{xx}{2}-3x+6\times \frac{x}{2}-18=2\left(\frac{x}{2}-1\right)^{2}$$

Cancel out 2, the greatest common factor in 6 and 2.

$$\frac{xx}{2}-3x+3x-18=2\left(\frac{x}{2}-1\right)^{2}$$

Combine -3x and 3x to get 0.

$$\frac{xx}{2}-18=2\left(\frac{x}{2}-1\right)^{2}$$

Multiply x and x to get x^{2}.

$$\frac{x^{2}}{2}-18=2\left(\frac{x}{2}-1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{x}{2}-1\right)^{2}.

$$\frac{x^{2}}{2}-18=2\left(\left(\frac{x}{2}\right)^{2}-2\times \frac{x}{2}+1\right)$$

To raise \frac{x}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{x^{2}}{2}-18=2\left(\frac{x^{2}}{2^{2}}-2\times \frac{x}{2}+1\right)$$

Express -2\times \frac{x}{2} as a single fraction.

$$\frac{x^{2}}{2}-18=2\left(\frac{x^{2}}{2^{2}}+\frac{-2x}{2}+1\right)$$

Cancel out 2 and 2.

$$\frac{x^{2}}{2}-18=2\left(\frac{x^{2}}{2^{2}}-x+1\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x+1 times \frac{2^{2}}{2^{2}}.

$$\frac{x^{2}}{2}-18=2\left(\frac{x^{2}}{2^{2}}+\frac{\left(-x+1\right)\times 2^{2}}{2^{2}}\right)$$

Since \frac{x^{2}}{2^{2}} and \frac{\left(-x+1\right)\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{x^{2}}{2}-18=2\times \frac{x^{2}+\left(-x+1\right)\times 2^{2}}{2^{2}}$$

Do the multiplications in x^{2}+\left(-x+1\right)\times 2^{2}.

$$\frac{x^{2}}{2}-18=2\times \frac{x^{2}-4x+4}{2^{2}}$$

Express 2\times \frac{x^{2}-4x+4}{2^{2}} as a single fraction.

$$\frac{x^{2}}{2}-18=\frac{2\left(x^{2}-4x+4\right)}{2^{2}}$$

Cancel out 2 in both numerator and denominator.

$$\frac{x^{2}}{2}-18=\frac{x^{2}-4x+4}{2}$$

Divide each term of x^{2}-4x+4 by 2 to get \frac{1}{2}x^{2}-2x+2.

$$\frac{x^{2}}{2}-18=\frac{1}{2}x^{2}-2x+2$$

Subtract \frac{1}{2}x^{2} from both sides.

$$\frac{x^{2}}{2}-18-\frac{1}{2}x^{2}=-2x+2$$

Combine \frac{x^{2}}{2} and -\frac{1}{2}x^{2} to get 0.

$$-18=-2x+2$$

Swap sides so that all variable terms are on the left hand side.

$$-2x+2=-18$$

Subtract 2 from both sides.

$$-2x=-18-2$$

Subtract 2 from -18 to get -20.

$$-2x=-20$$

Divide both sides by -2.

$$x=\frac{-20}{-2}$$

Divide -20 by -2 to get 10.

$$x=10$$