Найдите корни уравнения: (x-2)^2+24=(2+3x)^2.

\[(x - 2)^{2} + 24 = (2 + 3x)²\]

\[x^{2} - 4x + 4 + 24 =\]

\[= 4 + 12x + 9x^{2}\]

\[9x^{2} + 12x + 4 - x^{2} + 4x - 28 =\]

\[= 0\]

\[8x² + 16x - 24 = 0\ \ \ \ \ |\ :8\]

\[x² + 2x - 3 = 0\]

\[D = b^{2} - 4ac =\]

\[= 4 - 4 \cdot 1 \cdot ( - 3) = 4 + 12 = 16\]

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

\[x_{2} = \frac{- 2 - 4}{2} = - \frac{6}{2} = - 3\]

\[Ответ:x_{1} = 1;\ \ x_{2} = - 3.\]