Solve x^2+6x+1=(x+1)^2

$$x ^ { 2 } + 6 x + 1 = ( x + 1 ) ^ { 2 }$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

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

$$x^{2}+6x+1=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$x^{2}+6x+1-x^{2}=2x+1$$

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

$$6x+1=2x+1$$

Subtract 2x from both sides.

$$6x+1-2x=1$$

Combine 6x and -2x to get 4x.

$$4x+1=1$$

Subtract 1 from both sides.

$$4x=1-1$$

Subtract 1 from 1 to get 0.

$$4x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since 4 is not equal to 0, x must be equal to 0.

$$x=0$$