Найдите корни уравнения: (x-2)^2=3x-8.

\[(x - 2)^{2} = 3x - 8\]

\[x^{2} - 4x + 4 - 3x + 8 = 0\]

\[x^{2} - 7x + 12 = 0\]

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

\[= 49 - 4 \cdot 1 \cdot 12 = 49 - 48 = 1\]

\[x_{1} = \frac{7 + 1}{2} = \frac{8}{2} = 4\]

\[x_{2} = \frac{7 - 1}{2} = \frac{6}{2} = 3\]

\[Ответ:x = 4\ \ и\ \ \ x = 3.\]