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

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

Solve for x

$$x=5$$

Steps for Solving Linear Equation

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

$$x^{2}-4x-5=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$x^{2}-4x-5-x^{2}=-10x+25$$

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

$$-4x-5=-10x+25$$

Add 10x to both sides.

$$-4x-5+10x=25$$

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

$$6x-5=25$$

Add 5 to both sides.

$$6x=25+5$$

Add 25 and 5 to get 30.

$$6x=30$$

Divide both sides by 6.

$$x=\frac{30}{6}$$

Divide 30 by 6 to get 5.

$$x=5$$