We need to find the solution to the radical equation $\sqrt{2m+1} + m = 7$.

1. Problem Description

We need to find the solution to the radical equation $\sqrt{2m+1} + m = 7$.

2. Solution Steps

First, isolate the radical term:

$\sqrt{2m+1} = 7 - m$

Next, square both sides of the equation:

$(\sqrt{2m+1})^2 = (7-m)^2$

$2m+1 = 49 - 14m + m^2$

Rearrange the equation into a quadratic equation:

$0 = m^2 - 14m - 2m + 49 - 1$

$0 = m^2 - 16m + 48$

Factor the quadratic equation:

$0 = (m-4)(m-12)$

Solve for $m$:

$m-4=0$ or $m-12=0$

$m=4$ or $m=12$

Now, we need to check for extraneous solutions by substituting each value of $m$ back into the original equation $\sqrt{2m+1} + m = 7$.

For $m=4$:

$\sqrt{2(4)+1} + 4 = \sqrt{9} + 4 = 3 + 4 = 7$.

So, $m=4$ is a valid solution.

For $m=12$:

$\sqrt{2(12)+1} + 12 = \sqrt{25} + 12 = 5 + 12 = 17$.

Since $17 \neq 7$, $m=12$ is an extraneous solution.

3. Final Answer

The solution to the radical equation is $m=4$.