Calculus (3rd Edition)

There are parallel tangent lines at $x=-1, x=\frac{5}{3}$.
To have parallel tangent lines, we must have equal slopes; that is, $$\frac{dy}{dx}=3x^2, \quad \frac{dy}{dx}=2x+5,$$ then $3x^2=2x+5$. We get the equation $$3x^2-2x-5=0\Longrightarrow (3x-5)(x+1)=0.$$ There are parallel tangent lines at $x=-1, x=\frac{5}{3}$.