Make It An Ideal Expression

We have the equation $\large \frac{2x^2+5x+3}{3} = k^2$ for some integer $k$ . This gives us

$2x^2 + 5x + 3 = 3k^2 \implies 2x^2+5x+3-3k^2 = 0$ .

Now, we solve for $x$ using the quadratic formula in terms of $k$ , this gives us

$\large x = \frac{-5 + \sqrt{1+24k^2}}{4}$

Where the $\pm$ part of the equation was changed into $+$ , since the restriction that $x$ is natural forces us to search for positive values of $x$ .

Now, we need the discriminant to be a perfect square in order for $x$ to be rational, i.e.

$1+24k^2 = j^2 \implies j^2 - 24k^2 = 1$

for some integer $j$ .

It is easy to see that $(j,k) \in (5,1)$ satisfies, but unfortunately this yields $x = 0$ , but $0$ is not a natural number, so we are forced to look for more solutions.

Luckily, the Fibonacci-Brahmagupta identity comes to the rescue,

$(a^2-nb^2)(c^2 - nd^2) = (ac+nbd)^2 - n(ad+bc)^2$

We substitute $a = c = 5$ , $b = d = 1$ and $n = 24$ to get

$(5^2-(24)(1^2))(5^2-(24)(1^2)) = ((5)(5) + (24)(1)(1))^2 - (24)((5)(1) + (5)(1))^2$

Giving us $49^2 - 24(10)^2 = 1$ .

So, $(j,k) \in (49,10)$ also satisfies. This gives us a solution for $x$ :

$\large x = \frac{-5 + \sqrt{1+24k^2}}{4} = \frac{-5 + \sqrt{j^2}}{4} = \frac{j-5}{4}$

Since $1+24k^2 = j^2$ .

Substituting $j = 49$ gives us

$x = \frac{j-5}{4} = \frac{49-5}{4} = \frac{44}{4} = 11$

So our final answer is $\boxed{11}$ .