The number of real solution(s) of the equation $$x^2 + 3x + 2 = \min\{|x-3|,|x+2|\}\text{ is:}$$

To solve this systematically, let's break down the Right Hand Side (RHS) first.

The function $$\min\{|x - 3|, |x + 2|\}$$ gives the smaller of the two distances from $$x$$ to $$3$$ and from $$x$$ to $$-2$$. The point where these distances are exactly equal is the midpoint:

$$x = \frac{3 + (-2)}{2} = 0.5$$

This allows us to redefine the RHS piecewise:

• For $$x \le 0.5$$, the distance to $$-2$$ is smaller, so $$\text{RHS} = |x + 2|$$
• For $$x > 0.5$$, the distance to $$3$$ is smaller, so $$\text{RHS} = |x - 3|$$

Now, split the original equation into these two main regions and solve.

Region 1: For $$x \le 0.5$$

The equation becomes: $$x^2 + 3x + 2 = |x + 2|$$

Since the absolute value changes behavior at $$x = -2$$, we check two sub-cases:

• Case 1a ($$x \le -2$$):

  $$x^2 + 3x + 2 = -(x + 2)$$

  $$x^2 + 4x + 4 = 0$$

  $$(x + 2)^2 = 0 \implies x = -2$$

  (Since $$-2 \le -2$$, this is a valid solution).

• Case 1b ($$-2 < x \le 0.5$$):

  $$x^2 + 3x + 2 = +(x + 2)$$

  $$x^2 + 2x = 0$$

  $$x(x + 2) = 0 \implies x = 0 \text{ or } x = -2$$

  (Since $$x = 0$$ falls within $$-2 < x \le 0.5$$, this is a valid solution).

Region 2: For $$x > 0.5$$

The equation becomes: $$x^2 + 3x + 2 = |x - 3|$$

Again, split at the critical point $$x = 3$$:

• Case 2a ($$0.5 < x < 3$$):

  $$x^2 + 3x + 2 = -(x - 3)$$

  $$x^2 + 4x - 1 = 0$$

  Using the quadratic formula: $$x = \frac{-4 \pm \sqrt{16 - 4(1)(-1)}}{2} = -2 \pm \sqrt{5}$$

  Since $$\sqrt{5} \approx 2.23$$, the roots are roughly $$0.23$$ and $$-4.23$$. Neither of these falls in the interval $$(0.5, 3)$$.

  (No solution here).

• Case 2b ($$x \ge 3$$):

  $$x^2 + 3x + 2 = +(x - 3)$$

  $$x^2 + 2x + 5 = 0$$

  Checking the discriminant: $$D = 2^2 - 4(1)(5) = -16$$

  Since $$D < 0$$, there are no real roots.

  (No solution here).

Conclusion:

The only valid real solutions are $$x = -2$$ and $$x = 0$$.

Therefore, the number of real solutions is 2.