The number of the real solutions of the equation:
$$x|x+3|+|x-1|-2=0$$ is

Given:

$$x|x+3|+|x-1|-2=0$$

Case 1: $$ x < -3$$

If $$x < -3$$ then $$|x+3| = -(x + 3)$$ and $$|x -1| = -(x -1)$$

$$\Rightarrow$$ $$-x(x + 3) -(x-1) - 2 = 0$$

$$\Rightarrow$$ $$ -x^2 -3x - x + 1 - 2 = 0$$

$$\Rightarrow$$ $$ x^2 + 4x + 1 = 0$$

$$\Rightarrow$$ $$x^2 + 4x + 4 = -1 + 4$$

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

$$\Rightarrow$$ $$ (x + 2) = \sqrt{3} $$ OR $$(x +2) = -\sqrt{3}$$

$$\Rightarrow$$ $$ x = - 2 + \sqrt{3} \approx -0.268$$ OR $$x = -2 -\sqrt{3} \approx -3.732$$

But $$x < -3$$ $$\Rightarrow$$ $$x = -2 -\sqrt{3}$$ $$\Rightarrow$$ One solution from this case

Case 2: $$ -3 \leq x < 1$$

If $$ -3 \geq x < 1$$ then $$|x+3| = (x + 3)$$ and $$|x -1| = -(x -1)$$

$$\Rightarrow$$ $$x(x+3) -(x -1) - 2 = 0$$

$$\Rightarrow$$ $$x^2 + 2x - 1 = 0$$

$$\Rightarrow$$ $$(x + 1)^2 = 2$$

$$\Rightarrow$$ $$(x +1) = \sqrt{2}$$ OR $$(x + 1) = -\sqrt{2}$$

$$\Rightarrow$$ $$ x = -1 + \sqrt{2} \approx 0.414$$ OR $$x = -1 - \sqrt{2} \approx -2.414$$

Both of these satisfy the given condition

$$\Rightarrow$$ Two solutions from this case

Case 3: $$ x \geq 1$$

If $$x \geq 1$$ then $$|x+3| = (x + 3)$$ and $$|x -1| = (x -1)$$

$$\Rightarrow$$ $$ x(x +3) + (x -1) - 2 = 0$$

$$\Rightarrow$$ $$x^2 + 4x = 3$$

$$\Rightarrow$$ $$x^2 + 4x + 4 = 7$$

$$\Rightarrow$$ $$ (x + 2)^2 = 7$$

$$\Rightarrow$$ $$(x + 2) = \sqrt{7}$$ OR $$(x + 2) = -\sqrt{7}$$

$$\Rightarrow$$ $$x = -2 + \sqrt{7} \approx 0.646$$ OR $$x = -2 -\sqrt{7} \approx -4.646$$

None of these satisfies the condition. Thus zero solutions from this case.

Hence there are total of 3 solutions

$$x = -2 -\sqrt{3}, x = -1 + \sqrt{2}, x = -1 - \sqrt{2}$$

Hence, option C is the correct choice.