Solve x^2+b^2=(10-x)^2

$$x ^ { 2 } + b ^ { 2 } = ( 10 - x ) ^ { 2 }$$

Solve for x

$$x=-\frac{b^{2}}{20}+5$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-x\right)^{2}.

$$x^{2}+b^{2}=100-20x+x^{2}$$

Add 20x to both sides.

$$x^{2}+b^{2}+20x=100+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+b^{2}+20x-x^{2}=100$$

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

$$b^{2}+20x=100$$

Subtract b^{2} from both sides.

$$20x=100-b^{2}$$

Divide both sides by 20.

$$\frac{20x}{20}=\frac{100-b^{2}}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x=\frac{100-b^{2}}{20}$$

Divide -b^{2}+100 by 20.

$$x=-\frac{b^{2}}{20}+5$$

Solve for b (complex solution)

$$b=-2\sqrt{25-5x}$$

$$b=2\sqrt{25-5x}$$

Solve for b

$$b=2\sqrt{25-5x}$$

$$b=-2\sqrt{25-5x}\text{, }x\leq 5$$