Question
$$x^2+1 = (x+1)^2-2x$$
Answer
The equation has an infinite number of solutions.
Explanation
$$ \begin{aligned} x^2+1 &= (x+1)^2-2x&& \text{simplify right side} \\[1 em]x^2+1 &= x^2+2x+1-2x&& \\[1 em]x^2+1 &= x^2+2x+1-2x&& \\[1 em]x^2+1 &= x^2+1&& \text{move all terms to the left hand side } \\[1 em]x^2+1-x^2-1 &= 0&& \text{simplify left side} \\[1 em]x^2+1-x^2-1 &= 0&& \\[1 em]0 &= 0&& \\[1 em] \end{aligned} $$
Since the statement $ \color{blue}{ 0 = 0 } $ is TRUE for any value of $ x $, we conclude that the equation has infinitely many solutions.
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