Solve (x-3)^2=4^2+(x-5)^2

$$( x - 3 ) ^ { 2 } = 4 ^ { 2 } + ( x - 5 ) ^ { 2 }$$

Solve for x

$$x=8$$

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}.

$$x^{2}-6x+9=4^{2}+\left(x-5\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$x^{2}-6x+9=16+\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-6x+9=16+x^{2}-10x+25$$

Add 16 and 25 to get 41.

$$x^{2}-6x+9=41+x^{2}-10x$$

Subtract x^{2} from both sides.

$$x^{2}-6x+9-x^{2}=41-10x$$

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

$$-6x+9=41-10x$$

Add 10x to both sides.

$$-6x+9+10x=41$$

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

$$4x+9=41$$

Subtract 9 from both sides.

$$4x=41-9$$

Subtract 9 from 41 to get 32.

$$4x=32$$

Divide both sides by 4.

$$x=\frac{32}{4}$$

Divide 32 by 4 to get 8.

$$x=8$$