Solve left(x-3right)^2=x^2+18

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

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}.

$$x^{2}-6x+9=x^{2}+18$$

Subtract x^{2} from both sides.

$$x^{2}-6x+9-x^{2}=18$$

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

$$-6x+9=18$$

Subtract 9 from both sides.

$$-6x=18-9$$

Subtract 9 from 18 to get 9.

$$-6x=9$$

Divide both sides by -6.

$$x=\frac{9}{-6}$$

Reduce the fraction \frac{9}{-6} to lowest terms by extracting and canceling out 3.

$$x=-\frac{3}{2}$$