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

Critical points: $x=-3$ and $x=1$.
Case $x < -3$:
$x^2+4x+1=0 \Rightarrow x=-2\pm\sqrt{3}$.
Only $x=-2-\sqrt{3} \approx -3.73$ lies in $(-\infty,-3)$. (1 solution)
Case $-3 \le x < 1$:
$x^2+2x-1=0 \Rightarrow x=-1\pm\sqrt{2}$.
Both $x=-1+\sqrt{2} \approx 0.41$ and $x=-1-\sqrt{2} \approx -2.41$ lie in $[-3,1)$. (2 solutions)
Case $x \ge 1$:
$x^2+4x-3=0 \Rightarrow x=-2\pm\sqrt{7}$.
Neither root $\ge 1$. (0 solutions)
Total number of real solutions $= 3$.