Solve b^2-(5-x)^2=5^2-x^2

$$b ^ { 2 } - ( 5 - x ) ^ { 2 } = 5 ^ { 2 } - x ^ { 2 }$$

Solve for x

$$x=-\frac{b^{2}}{10}+5$$

Steps for Solving Linear Equation

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

$$b^{2}-\left(25-10x+x^{2}\right)=5^{2}-x^{2}$$

To find the opposite of 25-10x+x^{2}, find the opposite of each term.

$$b^{2}-25+10x-x^{2}=5^{2}-x^{2}$$

Calculate 5 to the power of 2 and get 25.

$$b^{2}-25+10x-x^{2}=25-x^{2}$$

Add x^{2} to both sides.

$$b^{2}-25+10x-x^{2}+x^{2}=25$$

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

$$b^{2}-25+10x=25$$

Subtract b^{2} from both sides.

$$-25+10x=25-b^{2}$$

Add 25 to both sides.

$$10x=25-b^{2}+25$$

Add 25 and 25 to get 50.

$$10x=50-b^{2}$$

Divide both sides by 10.

$$\frac{10x}{10}=\frac{50-b^{2}}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x=\frac{50-b^{2}}{10}$$

Divide 50-b^{2} by 10.

$$x=-\frac{b^{2}}{10}+5$$

Solve for b (complex solution)

$$b=-\sqrt{50-10x}$$

$$b=\sqrt{50-10x}$$

Solve for b

$$b=\sqrt{50-10x}$$

$$b=-\sqrt{50-10x}\text{, }x\leq 5$$