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.
 A: 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 $9+6\sqrt 3$.
A: You were correct in recognizing that $6x+9 = 3(2x+3)$ but we don't have to put everything in terms of $2x+3$
If we let $y=2x+3$ we get $2x^2 -7x\sqrt y +3y$ and we might recognize this as a quadratic.  If we replace $w =\sqrt y = \sqrt{2x+3}$ we get
$2x^2 - 7xw + 3w^2=0$ and maybe we can factor $2x^2 - 7xw + 3w^2=0$.  Can we?
And... sure we  can  $2x^2 -7xw+3w^2=(2x -w)(x-3w) = 0$.
So either $2x = w=\sqrt{2x+3}$ or $x=3w=3\sqrt{2x+3}$
These give us $x \ge 0; 4x^2 -2x-3=0$ or $x\ge 0; x^2 -18x -27$ which have at most $4$ real roots we can check.
$4x^2 -2x -3=0;x\ge 0\implies x=\frac {2+\sqrt{4+48}}8=\frac 14 + \frac {\sqrt{13}}4$.   And $x^2-18x-27=0;x\ge 0\implies x =\frac {18+\sqrt{18^2+108}}{2}=9 + \sqrt{81+27}=9+3\sqrt{9+3}=9+6\sqrt 3$
So barring the inevitable arithmetic errors that I always make, the answer is $9+6\sqrt 3$.
