Question
$$x^2+1 = (3x+1)^2$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = -\dfrac{ 3 }{ 4 } \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} x^2+1 &= (3x+1)^2&& \text{simplify right side} \\[1 em]x^2+1 &= 9x^2+6x+1&& \text{move all terms to the left hand side } \\[1 em]x^2+1-9x^2-6x-1 &= 0&& \text{simplify left side} \\[1 em]x^2+1-9x^2-6x-1 &= 0&& \\[1 em]-8x^2-6x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ -8x^{2}-6x = 0 } $, first we need to factor our $ x $.
$$ -8x^{2}-6x = x \left( -8x-6 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The second root can be found by solving equation $ -8x-6 = 0$.
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