Split at the breakpoints $x=-3$ and $x=1$.

Case I ($x\ge 1$): $|x+3|=x+3$, $|x-1|=x-1$.

$x(x+3)+(x-1)-2=0\Rightarrow x^2+4x-3=0\Rightarrow x=-2\pm \sqrt{7}$.

Both roots are less than 1, so both rejected. 0 solutions.

Case II ($-3\le x<1$): $|x+3|=x+3$, $|x-1|=1-x$.

$x(x+3)+(1-x)-2=0\Rightarrow x^2+2x-1=0\Rightarrow x=-1\pm \sqrt{2}$.

Both $\approx 0.41$ and $\approx -2.41$ lie in $[-3,1)$. 2 solutions.

Case III ($x<-3$): $|x+3|=-(x+3)$, $|x-1|=1-x$.

$-x(x+3)+(1-x)-2=0\Rightarrow x^2+4x+1=0\Rightarrow x=-2\pm \sqrt{3}$.

Only $-2-\sqrt{3}\approx -3.73$ is less than $-3$; the other is rejected. 1 solution.

Total: 3 real solutions.