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

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

Solve for x

$$x=1$$

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}.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 3x from both sides.

$$-6x+9-3x=0$$

Combine -6x and -3x to get -9x.

$$-9x+9=0$$

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$-9x=-9$$

Divide both sides by -9.

$$x=\frac{-9}{-9}$$

Divide -9 by -9 to get 1.

$$x=1$$