There are purely geometric ways to do the problem, but since it arose in a calculus course, let's use standard max/min procedures.
Think of a "general" (unspecified) point $P$ on the line $y=2x+3$. Suppose that the $x$-coordinate of $P$ is $x$. Then the $y$-coordinate of $P$ is $2x+3$. So $P$ is the point $(x,2x+3)$.
What is the distance from $(x,2x+3)$ to the origin? By what I hope is a familiar formula, this distance is
$$\sqrt{(x-0)^2 +(2x+3-0)^2}.$$
This is because in general the distance between $(a,b)$ and $(c,d)$ is $\sqrt{(a-c)^2+(b-d)^2}$. Apply this formula, using $a=x$, $b=2x+3$, and $c=0$, $d=0$.
Alternately, you can find the distance from $(x,2x+3)$ to the origin by drawing a picture and using the Pythagorean Theorem.
We want to choose $x$ so as to minimize the distance from $P$ to the origin, so we want to minimize $\sqrt{x^2+(2x+3)^2}$.
We can now let
$$f(x)=\sqrt{x^2+(2x+3)^2}$$
and find the $x$ that makes $f(x)$ smallest by the usual derivative process.
But there is a little trick that simplifies things. If $\sqrt{x^2+(2x+3)^2}$ is as small as possible, then so is $x^2+(2x+3)^2$, and vice-versa. So we try to minimize the square of the distance.
Let
$$g(x)=x^2+(2x+3)^2.$$
Can you now find the value of $x$ that makes $g(x)$ as small as possible?
Added: After some simplification,
$$f'(x)= \frac{5x+6}{\left(x^2+(2x+3)^2\right)^{1/2}}.$$
If you take the suggested alternate route, you will find that
$$g'(x)=10x+12.$$
Best Answer
HINT: $d^2=x^2+\frac 4 x = x^2+ \frac 2 x+\frac 2 x $...