A linear first-order equation with polynomial right-hand side

Homogeneous equation. $y' + 2y = 0$ gives
$$y_h(x) = Ce^{-2x}.$$

Particular solution. Try $y_p = ax^2 + bx + c$. Then $y_p' = 2ax + b$, and
$$y_p' + 2y_p = 2ax + b + 2ax^2 + 2bx + 2c = 2ax^2 + (2a+2b)x + (b+2c).$$
Identifying with $x^2 - 2x + 3$:
$$2a = 1 \implies a = \frac12,$$
$$2a + 2b = -2 \implies 1 + 2b = -2 \implies b = -\frac32,$$
$$b + 2c = 3 \implies -\frac32 + 2c = 3 \implies c = \frac94.$$
So
$$y_p(x) = \frac12 x^2 - \frac32 x + \frac94.$$

General solution.
$$y(x) = Ce^{-2x} + \frac12 x^2 - \frac32 x + \frac94, \qquad C \in \mathbb{R}.$$