Given the function f(x) = 3+2x−x2, draw its graph for -2 ext and 4 and find the solution of f(x) = 0.

• For $x = -2$: $y = 3 + 2(-2) - (-2)^2 = 3 - 4 - 4 = -5$
• For $x = -1$: $y = 3 + 2(-1) - (-1)^2 = 3 - 2 - 1 = 0$
• For $x = 0$: $y = 3 + 2(0) - (0)^2 = 3 + 0 - 0 = 3$
• For $x = 1$: $y = 3 + 2(1) - (1)^2 = 3 + 2 - 1 = 4$
• For $x = 2$: $y = 3 + 2(2) - (2)^2 = 3 + 4 - 4 = 3$
• For $x = 3$: $y = 3 + 2(3) - (3)^2 = 3 + 6 - 9 = 0$
• For $x = 4$: $y = 3 + 2(4) - (4)^2 = 3 + 8 - 16 = -5$

The table of values is:

$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & -5 & 0 & 3 & 4 & 3 & 0 & -5 \\ \hline \end{array}$$

Plot these points on a graph paper using the specified scale ($2 \, \text{cm}$ for $1 \, \text{unit}$ on both axes) and draw a smooth curve through them.

From the table of values, we see that $y=0$ when $x=-1$ and $x=3$.
Analytically, we solve $3+2x-x^2 = 0$:

$$x^2 - 2x - 3 = 0$$ $$(x-3)(x+1) = 0$$ $$x-3 = 0 \quad \text{or} \quad x+1 = 0$$ $$x = 3 \quad \text{or} \quad x = -1$$

The solutions are $\boxed{x = -1, x = 3}$.