Answer
$x=3$
Work Step by Step
We are given:
$\sqrt{3\sqrt{2x+3}}=\sqrt{5x-6}$
We square both sides:
$(\sqrt{3\sqrt{2x+3}})^{2}=(\sqrt{5x-6})^{2}$
$3\sqrt{2x+3}=5x-6$
And square both sides again:
$(3\sqrt{2x+3})^{2}=(5x-6)^{2}$
$9\ (2x+3)=25x^{2}-60x+36$
And distribute:
$18x+27=25x^{2}-60x+36$
$25x^{2}-78x+9=0$
Factor the trinomial:
$(25x-3)(x-3)=0$
Use the zero-factor property by equating each factor to zero:
$(25x-3)=0$ or $(x-3)=0$
$x=\displaystyle \frac{3}{25}$ or $x=3$
However, the solution $x=\frac{3}{25}$ does not work in the original equation:
$\sqrt{3\sqrt{2(\frac{3}{25})+3}}=\sqrt{5*(\frac{3}{25})-6}$
$\sqrt{3\sqrt{\frac{6}{25}+3}}=\sqrt{\frac{3}{5}-6}$
$\sqrt{3\sqrt{\frac{6}{25}+\frac{75}{25}}}=\sqrt{\frac{3}{5}-\frac{30}{5}}$
$\sqrt{3\sqrt{\frac{81}{25}}}=\sqrt{-\frac{27}{5}}$
Since we got an untrue statement (the right side is imaginary), the solution
$x=\frac{3}{25}$ does not work.
Thus the only solution is $x=3$.