$x^2 - 3x - 5 = 0$
по теореме Виета:
$\begin{equation*} \begin{cases} x_1 + x_2 = -b &\\ x_1x_2 = c & \end{cases} \end{equation*}$
$\begin{equation*} \begin{cases} x_1 + x_2 = 3 &\\ x_1x_2 = -5 & \end{cases} \end{equation*}$
$y_1 = x_1 - 1$
$y_2 = x_2 - 1$
$y_1 + y_2 = x_1 - 1 + x_2 - 1 = (x_1 + x_2) - 2 = 3 - 2 = 1$
$y_1y_2 = (x_1 - 1)(x_2 - 1) = x_1x_2 - x_2 - x_1 + 1 = x_1x_2 - (x_1 + x_2) + 1 = -5 - 3 + 1 = -7$
тогда:
$y_1 + y_2 = -b_2$
$b_2 = -(y_1 + y_2)$
$b_2 = -1$
$y_1y_2 = c_2$
$c_2 = -7$
$ax^2 - b_2x + c_2 = 0$
тогда:
$x^2 - x - 7 = 0$
Ответ: $x^2 - x - 7 = 0$