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

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

Solve for x

$$x=\frac{1}{4}=0.25$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2 by x-3.

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

Use the distributive property to multiply 2x-6 by x-2 and combine like terms.

$$2x^{2}-10x+12-\left(x+3\right)^{2}=\left(x+1\right)\left(x-1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$2x^{2}-10x+12-\left(x^{2}+6x+9\right)=\left(x+1\right)\left(x-1\right)$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$2x^{2}-10x+12-x^{2}-6x-9=\left(x+1\right)\left(x-1\right)$$

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

$$x^{2}-10x+12-6x-9=\left(x+1\right)\left(x-1\right)$$

Combine -10x and -6x to get -16x.

$$x^{2}-16x+12-9=\left(x+1\right)\left(x-1\right)$$

Subtract 9 from 12 to get 3.

$$x^{2}-16x+3=\left(x+1\right)\left(x-1\right)$$

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.

$$x^{2}-16x+3=x^{2}-1$$

Subtract x^{2} from both sides.

$$x^{2}-16x+3-x^{2}=-1$$

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

$$-16x+3=-1$$

Subtract 3 from both sides.

$$-16x=-1-3$$

Subtract 3 from -1 to get -4.

$$-16x=-4$$

Divide both sides by -16.

$$x=\frac{-4}{-16}$$

Reduce the fraction \frac{-4}{-16} to lowest terms by extracting and canceling out -4.

$$x=\frac{1}{4}$$