Найдите корни уравнения: (x+3)^2-16=(1-2x)^2.

\[(x + 3)^{2} - 16 = (1 - 2x)^{2}\]

\[x^{2} + 6x + 9 - 16 =\]

\[= 1 - 4x + 4x^{2}\]

\[x^{2} + 6x - 7 - 1 + 4x - 4x^{2} = 0\]

\[- 3x^{2} + 10x - 8 = 0\ \ \ \ | \cdot ( - 1)\]

\[3x^{2} - 10x + 8 = 0\]

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

\[= 100 - 4 \cdot 3 \cdot 8 = 100 - 96 =\]

\[= 4\]

\[x_{1} = \frac{10 + 2}{6} = \frac{12}{6} = 2\]

\[x_{2} = \frac{10 - 2}{6} = \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3}\]

\[Ответ:x = 2\ \ и\ \ \ x = 1\frac{1}{3}.\]