Найдите корни уравнения: 4·(x-1)^2=12x+3.

\[4 \cdot (x - 1)^{2} = 12x + 3\]

\[4 \cdot \left( x^{2} - 2x + 1 \right) = 12x + 3\]

\[4x^{2} - 8x + 4 - 12x - 3 = 0\]

\[4x² - 20x + 1 = 0\ \]

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

\[= 400 - 4 \cdot 4 \cdot 1 =\]

\[= 400 - 16 = 384\]

\[x_{1} = \frac{20 + 8\sqrt{6}}{8} = 2,5 + \sqrt{6}\]

\[x_{2} = \frac{20 - 8\sqrt{6}}{8} = 2,5 - \sqrt{6}\]

\[Ответ:x_{1} = 2,5 + \sqrt{6};\ \ \]

\[x_{2} = 2,5 - \sqrt{6}.\]