To solve the equation \( 5^{2x + 1} = 125^x + 25 \), we'll start by rewriting \( 125 \) and \( 25 \) as powers of \( 5 \).

\[
125 = 5^3 \quad \text{and} \quad 25 = 5^2
\]

Now, substituting these into the equation, we get:

\[
5^{2x + 1} = (5^3)^x + 5^2
\]

This simplifies to:

\[
5^{2x + 1} = 5^{3x} + 5^2
\]

Next, we rewrite \( 5^{2x + 1} \) in terms of \( 5^{3x} \):

\[
5^{2x + 1} = 5^{2x} \cdot 5^1 = 5 \cdot 5^{2x}
\]

Now, the equation is:

\[
5 \cdot 5^{2x} = 5^{3x} + 5^2
\]

We can divide the entire equation by \( 5 \) (provided \( 5 \neq 0 \)):

\[
5^{2x} = 5^{3x - 1} + 5
\]

Now, substituting \( 5^{2x} = y \) (where \( y = 5^{2x} \)), our equation becomes:

\[
y = 5^{3x - 1} + 5
\]

Recall that \( 5^{3x - 1} = \frac{y^{3/2}}{5} \) since \( 5^{3x} = y^{3/2} \). Thus, we rewrite the equation:

\[
y = \frac{y^{3/2}}{5} + 5
\]

Now we multiply everything by \( 5 \) to eliminate the fraction:

\[
5y = y^{3/2} + 25
\]

Rearranging gives us:

\[
y^{3/2} - 5y + 25 = 0
\]

Now we can let \( z = y^{1/2} \), and then \( y^{3/2} = z^3 \) and \( y = z^2 \). Substituting this into the equation results in:

\[
z^3 - 5z^2 + 25 = 0
\]

Next, we can try to find the roots of \( z^3 - 5z^2 + 25 \). To apply the Rational Root Theorem or synthetic division, we can test for potential rational roots by using values like \( 1, 2, -1, -2, 5, \) etc.

However, upon testing these values, it might require numerical methods or further polynomial techniques, as directly factoring by inspection may not yield results quickly.

Instead, let's analyze the original power equation we had:

From \( 5^{2x + 1} = 125^x + 25 \), we can also take the logarithm:

Taking logs on both sides:

\[
\log(5^{2x + 1}) = \log(125^x + 25)
\]

This could lead to working towards a numerical estimate or solution.

Given the mathematical complexity, the final solution may require numerical calculation or approximation. You could also apply software or a graphing calculator to find intersections and roots of \( f(x) = 5^{2x + 1} − (125^x + 25) \) to find the value of \( x \) within an acceptable domain.

If you are seeking an exact algebraic solution, further iteration or more advanced root-finding techniques are helpful.

Final solutions can be approximated graphically or through softwares.