Question
$$1 = (x+2)^2-3$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 1 &= (x+2)^2-3&& \text{simplify right side} \\[1 em]1 &= x^2+4x+4-3&& \\[1 em]1 &= x^2+4x+1&& \text{move all terms to the left hand side } \\[1 em]1-x^2-4x-1 &= 0&& \text{simplify left side} \\[1 em]1-x^2-4x-1 &= 0&& \\[1 em]-x^2-4x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -x^{2}-4x = 0 } $, first we need to factor our $ x $.
$$ -x^{2}-4x = x \left( -x-4 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -x-4 = 0$.
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