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

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

Solve for x

$$x=4$$

Steps for Solving Linear Equation

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

$$x^{2}-6x+9=x^{2}-3\left(x+1\right)$$

Use the distributive property to multiply -3 by x+1.

$$x^{2}-6x+9=x^{2}-3x-3$$

Subtract x^{2} from both sides.

$$x^{2}-6x+9-x^{2}=-3x-3$$

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

$$-6x+9=-3x-3$$

Add 3x to both sides.

$$-6x+9+3x=-3$$

Combine -6x and 3x to get -3x.

$$-3x+9=-3$$

Subtract 9 from both sides.

$$-3x=-3-9$$

Subtract 9 from -3 to get -12.

$$-3x=-12$$

Divide both sides by -3.

$$x=\frac{-12}{-3}$$

Divide -12 by -3 to get 4.

$$x=4$$