Найдите множество решений неравенства: (x-4)^2+12>=(3x-2)^2.

\[(x - 4)^{2} + 12 \geq (3x - 2)²\]

\[- 8x^{2} + 4x + 24 \geq 0\ \ \ |\ :4\]

\[- 2x^{2} + x + 6 \geq 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{- 1 + 7}{- 4} = - 1,5\]

\[x_{2} = \frac{- 1 - 7}{- 4} = 2\]

\[Ответ:\lbrack - 1,5;2\rbrack.\]