$\left ( x + 1 \right )^{2} = 4\left ( x^{2} - 2x + 1 \right )^{2}$
$\Leftrightarrow \left | x + 1 \right | = \left | 2\left ( x^{2} - 2x + 1 \right ) \right |$
$\Leftrightarrow \left[ \begin{array}{l}x + 1 = 2\left ( x^{2} - 2x + 1 \right )\\x + 1 = -2\left ( x^{2} - 2x + 1 \right )\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x + 1 = 2x^{2} - 4x + 2\\x + 1 = -2x^{2} + 4x - 2\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}2x^{2} - 5x + 1 = 0\\-2x^{2} + 3x - 3 = 0\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{5 - \sqrt{17}}{4}\\x = \dfrac{5 + \sqrt{17}}{4}\end{array} \right.$
