The only point of inflection on the line representing the equation $$y=x^3+2x^2-5$$ is at:

Inflection points represent points on the graph where the curve changes from concave up to concave down or vice versa.

Mathematically, inflection points on the graph are equal to points where the 2nd derivative is equal to 0. The second derivative essentially gives us information regarding if the slope of the line tangent to the graph is increasing or decreasing. At inflection points, the slope of the line transitions from increasing to decreasing, or vice versa, and therefore the 2nd derivative must cross 0.

Our first step is to find the second derivative of the equation:

$$\begin{gathered} y=x^3+2x^2-5 \\ {y}^{\text{'}}=3x^2+4x \\ {y}^{\text{'}\text{'}}=6x+4 \end{gathered}$$

Now that we have the equation of our 2nd derivative, we can solve it to see at what x value it becomes 0:

$$\begin{gathered} y^{\text{'}\text{'}}=6x+4 \\ 0 = 6x+4 \\ -4=6x \\ x = \frac{-2}{3} \end{gathered}$$

Therefore, the only inflection point is at $x = \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$.  We can plug this value into our original equation to find the y value associated with this x value:

$$\begin{gathered} y=x^3+2x^2-5 \\ y = {\left(\frac{-2}{3}\right)}^3+2{\left(\frac{-2}{3}\right)}^2-5 \\ y = \left(\frac{-8}{27}\right)+2\left(\frac{4}{9}\right)-5 \\ y = \frac{-8}{27}+\frac{24}{27}-\frac{135}{27}=\frac{-119}{27} \end{gathered}$$

Therefore, our only point of inflection is at $\mathbf{(}\frac{\mathbf{-}\mathbf{2}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{119}}{\mathbf{27}}\mathbf{)}$We can visualize this concave down to concave up inflection point on the graph: