Notice that when [tex]x=-1[/tex], you have
[tex](-1)^3+2(-1)+3=-1-2+3=0[/tex]
so [tex]x=-1[/tex] is a solution, and the cubic has [tex]x+1[/tex] as a factor. This means you can divide to get another quadratic factor and thus two more solutions.
[tex]\dfrac{x^3+2x+3}{x+1}=x^2-x+3[/tex]
So,
[tex]x^3+2x+3=(x+1)(x^2-x+3)=0[/tex]
You can use the quadratic formula here:
[tex]x=\dfrac{1\pm\sqrt{1-12}}2=\dfrac12\pm i\dfrac{\sqrt{11}}2[/tex]