To find the point on the line $y = 2x + 3$ that is closest to the origin, we need to minimize the distance from a point $(x, y)$ on the line to the origin $(0, 0)$. The distance $d$ from the origin to a point $(x, y)$ is given by:

$d = \sqrt{x^2 + y^2}$

Since $y = 2x + 3$, we can substitute for $y$ in the distance formula:

$d = \sqrt{x^2 + (2x + 3)^2}$

To minimize $d$, it suffices to minimize $d^2 = x^2 + (2x + 3)^2$ because the square root is a monotonically increasing function.

Expand $(2x + 3)^2$:

$(2x + 3)^2 = 4x^2 + 12x + 9$

Thus, $d^2$ becomes:

$d^2 = x^2 + 4x^2 + 12x + 9 = 5x^2 + 12x + 9$

To find the minimum, take the derivative of $d^2$ with respect to $x$ and set it to zero:

$\frac{d}{dx}(5x^2 + 12x + 9) = 10x + 12$

Set the derivative equal to zero:

$10x + 12 = 0$

Solve for $x$:

x = -\frac{6}{5}$$ Substitute $$x = -\frac{6}{5}$$ back into the line equation to find $$y$$: $$y = 2\left(-\frac{6}{5}\right) + 3 = -\frac{12}{5} + \frac{15}{5} = \frac{3}{5}$$ Therefore, the point on the line $$y = 2x + 3$$ that is closest to the origin is: $$\left(-\frac{6}{5}, \frac{3}{5}\right)$$