\[ (x+2)^2 = x^2 + 2 \cdot x \cdot 2 + 2^2 = x^2 + 4x + 4 \]
\[ x^2 + 4x + 4 = 4x^2 + 4x - 32 \]
\[ x^2 + 4x + 4 - 4x^2 - 4x + 32 = 0 \]
\[ -3x^2 + 36 = 0 \]
\[ x^2 - 12 = 0 \]
\[ x^2 = 12 \]
\[ x = \pm \sqrt{12} = \pm 2\sqrt{3} \]
Ответ: ±2√3