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Let's solve the equation step by step:

The given equation is:

\[
\sqrt{10x + 50} = x + 5
\]

### Step 1: Square both sides
Squaring both sides of the equation will eliminate the square root:

\[
(\sqrt{10x + 50})^2 = (x + 5)^2
\]

This simplifies to:

\[
10x + 50 = (x + 5)(x + 5)
\]

### Step 2: Expand the right-hand side
Now, expand the right side:

\[
(x + 5)(x + 5) = x^2 + 10x + 25
\]

Thus, the equation becomes:

\[
10x + 50 = x^2 + 10x + 25
\]

### Step 3: Simplify the equation
Subtract \(10x + 50\) from both sides:

\[
0 = x^2 + 10x + 25 - 10x - 50
\]

Simplifying further:

\[
0 = x^2 - 25
\]

### Step 4: Solve for \(x\)
Now, solve for \(x\) by factoring the equation:

\[
x^2 - 25 = (x - 5)(x + 5) = 0
\]

So, \(x = 5\) or \(x = -5\).

### Step 5: Verify the solutions
We need to check for extraneous solutions by substituting these values back into the original equation.

- For \(x = 5\):

\[
\sqrt{10(5) + 50} = 5 + 5
\]

\[
\sqrt{100} = 10 \quad \text{which is true.}
\]

- For \(x = -5\):

\[
\sqrt{10(-5) + 50} = -5 + 5
\]

\[
\sqrt{-50 + 50} = 0 = 0 \quad \text{which is also true.}
\]

### Conclusion:
Both \(x = 5\) and \(x = -5\) are valid solutions.

Thus, the correct answer is:

\[
\boxed{x = 5 \text{ or } x = -5}
\]

Let me know if you'd like more details or have questions! Here are some related questions:

1. What are the conditions under which a square root equation could have no solution?
2. How would the solution change if the equation had been \( \sqrt{10x + 50} = x + 7 \)?
3. What happens if you square both sides of an equation and it introduces an extraneous solution?
4. How can you quickly check whether a solution is extraneous after solving the equation?
5. What types of equations are best solved by squaring both sides?

**Tip:** Always check for extraneous solutions when squaring both sides of an equation!

Learn how to solve the radical equation √(10x + 50) = x + 5. This problem involves squaring both sides and factoring to find the solutions. Suitable for students in Grades 9-10.

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