\[\boxed{\mathbf{734\ (734).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[x^{2} - 12x + 4 = 0\ \]
\[x_{1} + x_{2} = 12;\ \ \ \ x_{1} \cdot x_{2} = 4\ \]
\[Корни\ нового\ уравнения\ \]
\[на\ 3\ больше\ данных:\]
\[y_{1} \cdot y_{2} = \left( x_{1} + 3 \right)\left( x_{2} + 3 \right) =\]
\[= x_{1}x_{2} + 3x_{1} + 3x_{2} + 9 =\]
\[= x_{1}x_{2} + 3 \cdot \left( x_{1} + x_{2} \right) + 9 =\]
\[= 4 + 3 \cdot 12 + 9 = 49\]
\[y_{1} + y_{2} = x_{1} + 3 + x_{2} + 3 =\]
\[= \left( x_{1} + x_{2} \right) + 6 = 12 + 6 = 18\ \]
\[Новое\ уравнение:\]
\[y^{2} - 18y + 49 = 0\]