Solve (x-2)^2-(x+1)(x-1)=5

$$( x - 2 ) ^ { 2 } - ( x + 1 ) ( x - 1 ) = 5$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

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

$$x^{2}-4x+4-\left(x+1\right)\left(x-1\right)=5$$

Consider \left(x+1\right)\left(x-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$x^{2}-4x+4-\left(x^{2}-1\right)=5$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}-4x+4-x^{2}+1=5$$

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

$$-4x+4+1=5$$

Add 4 and 1 to get 5.

$$-4x+5=5$$

Subtract 5 from both sides.

$$-4x=5-5$$

Subtract 5 from 5 to get 0.

$$-4x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -4 is not equal to 0, x must be equal to 0.

$$x=0$$