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

Let's say: $$g(x) = \min\{|x-3|, |x+2|\}$$

Equating the two terms:

$$|x-3| = |x+2| \implies x-3 = -(x+2) \implies 2x = 1 \implies x = 0.5$$

For $$x \le 0.5$$, the distance to $$-2$$ is smaller or equal to the distance to $$3$$, so $$g(x) = |x+2|$$

If $$x \le -2$$, $$g(x) = -(x+2)$$

If $$-2 < x \le 0.5$$, $$g(x) = x+2$$

For $$x > 0.5$$, the distance to $$3$$ is smaller, so $$g(x) = |x-3|$$

If $$0.5 < x \le 3$$, $$g(x) = 3-x$$

If $$x > 3$$, $$g(x) = x-3$$

Now, solving $$f(x) = g(x)$$ in each interval

Interval 1: $$x \le -2$$

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

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

$$(x+2)^2=0\Rightarrow\mathbf{x=-2}$$

Since $$-2$$ is in the interval, this is the first solution.

Interval 2: $$-2 < x \le 0.5$$

$$x^2+3x+2=x+2$$

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

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

$$x = 0$$ is in the interval $$(-2, 0.5]$$, so $$\mathbf{x = 0}$$ is the second solution.

$$x=-2$$ is already found.

Interval 3: $$0.5 < x \le 3$$

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

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

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

$$x \approx -2 + 2.236 = 0.236$$ (Not in the interval $$(0.5, 3]$$)

$$x \approx -2 - 2.236 = -4.236$$ (Not in the interval $$(0.5, 3]$$)

So, no solutions in this interval.

Interval 4: $$x > 3$$

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

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

$$D=2^2-4(1)(5)=4-20=-16$$

Since D < 0, there are no real solutions.

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

Hence, the number of real solutions is 2.