Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$ 
Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$

Source: https://www.hkage.edu.hk/uploads/file/202207/6cda89c718b674f6ac3aa2c19049abe5.pdf
I tried substituting $y = 2x+3$ and ended up with $\frac{y^2}{2}-\frac{7}{2}y^{\frac 3 2} + \frac {21}{2} y^{\frac 1 2} + \frac 9 2 = 0$ but I'm still not sure how to find the largest $y$ from this.
I also tried calculating the derivative, ending up with $(4x+6)\sqrt{2x+3}-21x-21=0$. However, solving for $x$ is not trivial after simplifying the equation.
Also, I believe that calculus is not needed in this contest so I am looking for an elementary solution to this problem. Thanks in advance.

Notice that
\begin{align*}
2x^2+6x+9-7x\sqrt{2x+3}&=2x^2-7x\sqrt{2x+3}+3(2x+3)\\
&=\left(x-3\sqrt{2x+3}\right)\left(2x-\sqrt{2x+3}\right).
\end{align*}
The real solution of $x-3\sqrt{2x+3}=0$ is $x=9+6\sqrt 3$ and the real solution of $2x-\sqrt{2x+3}=0$ is $x=\frac{1+\sqrt{13}}4$. Finally,
$$9+6\sqrt 3>9>\frac{1+4}4>\frac{1+\sqrt{13}}4.$$
Therefore, the desired answer is $9+6\sqrt 3$.
