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