Solve left(x-3right)^2=4x+x^2

$${ \left(x-3 \right) }^{ 2 } = 4x+ { x }^{ 2 }$$

Solve for x

$$x=\frac{9}{10}=0.9$$

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}.

$$x^{2}-6x+9=4x+x^{2}$$

Subtract 4x from both sides.

$$x^{2}-6x+9-4x=x^{2}$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x+9=x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}-10x+9-x^{2}=0$$

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

$$-10x+9=0$$

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

$$-10x=-9$$

Divide both sides by -10.

$$x=\frac{-9}{-10}$$

Fraction \frac{-9}{-10} can be simplified to \frac{9}{10} by removing the negative sign from both the numerator and the denominator.

$$x=\frac{9}{10}$$