Solve (x-5)(x+5)-left(x-1right)^2=0

$$(x-5)(x+5)- { \left(x-1 \right) }^{ 2 } =0$$

Solve for x

$$x=13$$

Steps for Solving Linear Equation

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$x^{2}-25-\left(x-1\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$x^{2}-25-\left(x^{2}-2x+1\right)=0$$

To find the opposite of x^{2}-2x+1, find the opposite of each term.

$$x^{2}-25-x^{2}+2x-1=0$$

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

$$-25+2x-1=0$$

Subtract 1 from -25 to get -26.

$$-26+2x=0$$

Add 26 to both sides. Anything plus zero gives itself.

$$2x=26$$

Divide both sides by 2.

$$x=\frac{26}{2}$$

Divide 26 by 2 to get 13.

$$x=13$$