Solve the Equation: Finding x in (x+1)² = x² + 13

$(x+1)^2=x^2+13$

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$We'll apply the mentioned formula and expand the parentheses in the expressions in the equation:

$(x+1)^2=x^2+13 \\ x^2+2\cdot x\cdot1+1^2=x^2+13 \\ x^2+2x+1=x^2+13$

We'll continue and combine like terms, by moving terms between sides. Then we can notice that the squared term cancels out, therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2+2x+1=x^2+13 \\ 2x=12\hspace{8pt}\text{/}:2\\ \boxed{x=6}$Therefore, the correct answer is answer B.