To solve the equation
\[ \sqrt{6x + 40} = x + 8, \]
we will first square both sides to eliminate the square root.
\[ (\sqrt{6x + 40})^2 = (x + 8)^2. \]
This simplifies to:
\[ 6x + 40 = (x + 8)(x + 8). \]
Now expand the right side:
\[ 6x + 40 = x^2 + 16x + 64. \]
Next, we will rearrange the equation to set it to zero:
\[ 0 = x^2 + 16x + 64 - 6x - 40. \]
Combine like terms:
\[ 0 = x^2 + 10x + 24. \]
Now we have a quadratic equation:
\[ x^2 + 10x + 24 = 0. \]
We can factor this equation:
\[ (x + 4)(x + 6) = 0. \]
Setting each factor equal to zero gives us the possible solutions:
- \(x + 4 = 0 \implies x = -4\)
- \(x + 6 = 0 \implies x = -6\)
Now we have two possible solutions: \(x = -4\) and \(x = -6\).
However, we need to check each solution in the original equation to make sure they are valid (since squaring can introduce extraneous solutions).
- Check \(x = -4\):
\[ \sqrt{6(-4) + 40} = -4 + 8 \implies \sqrt{-24 + 40} = 4 \implies \sqrt{16} = 4, \] which is true.
- Check \(x = -6\):
\[ \sqrt{6(-6) + 40} = -6 + 8 \implies \sqrt{-36 + 40} = 2 \implies \sqrt{4} = 2, \] which is also true.
Both solutions are valid. Thus, the final solutions are:
\[ \boxed{-4 \text{ and } -6}. \]