To solve the equation

\[
\frac{2x + 1}{x + 9} = 5x,
\]

we first want to cross-multiply to eliminate the fraction. This gives us:

\[
2x + 1 = 5x(x + 9).
\]

Now, expanding the right side:

\[
2x + 1 = 5x^2 + 45x.
\]

Next, we move all terms to one side of the equation:

\[
0 = 5x^2 + 45x - 2x - 1 \\
0 = 5x^2 + 43x - 1.
\]

This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 5 \), \( b = 43 \), and \( c = -1 \).

We can solve for \( x \) using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
\]

Calculating the discriminant:

\[
b^2 - 4ac = 43^2 - 4 \cdot 5 \cdot (-1) \\
= 1849 + 20 = 1869.
\]

Now substituting into the quadratic formula:

\[
x = \frac{-43 \pm \sqrt{1869}}{2 \cdot 5} \\
= \frac{-43 \pm \sqrt{1869}}{10}.
\]

Now, we can simplify \(\sqrt{1869}\) if possible. However, \(1869\) can be factored as \(3 \times 623\), where \(623\) is a prime number, hence it cannot be simplified further.

Thus the final solutions for \(x\) are:

\[
x = \frac{-43 + \sqrt{1869}}{10} \quad \text{and} \quad x = \frac{-43 - \sqrt{1869}}{10}.
\]

This gives us the two roots of the quadratic equation.