Rewrite the exponent on the left side of the equation.

The given equation is:

$$5^{2x+6} = 7^{x+3}$$

Notice that $2x+6$ can be factored as $2(x+3)$.
So, the equation becomes:

$$5^{2(x+3)} = 7^{x+3}$$

$$(5^2)^{x+3} = 7^{x+3}$$ $$25^{x+3} = 7^{x+3}$$

Divide both sides by $7^{x+3}$:

$$\frac{25^{x+3}}{7^{x+3}} = 1$$

For any non-zero base $B$, if $B^A = 1$, then the exponent $A$ must be 0.
In this case, $B = \frac{25}{7}$ and $A = x+3$.
Therefore, we must have:

$$x+3 = 0$$ $$x = -3$$

Substitute $x=-3$ back into the original equation:
Left side: $5^{2(-3)+6} = 5^{-6+6} = 5^0 = 1$
Right side: $7^{(-3)+3} = 7^0 = 1$
Since $1=1$, the solution is correct.

The final answer is $-3$.