To solve the equation
\[ \sqrt{x + 4} + \sqrt{3x + 16} = 2, \]
we start by isolating one of the square root terms. Let's isolate \(\sqrt{x + 4}\):
\[ \sqrt{x + 4} = 2 - \sqrt{3x + 16}. \]
Next, we square both sides to eliminate the square root:
\[ x + 4 = (2 - \sqrt{3x + 16})^2. \]
Expanding the right-hand side, we have:
\[ x + 4 = 4 - 4\sqrt{3x + 16} + (3x + 16). \]
Combining like terms on the right gives:
\[ x + 4 = 3x + 20 - 4\sqrt{3x + 16}. \]
Now, let's move all terms involving \(x\) to one side:
\[ x + 4 - 3x - 20 = -4\sqrt{3x + 16}. \]
Simplifying this gives:
\[ -2x - 16 = -4\sqrt{3x + 16}. \]
Now, divide both sides by \(-4\):
\[ \frac{2x + 16}{4} = \sqrt{3x + 16}. \]
This simplifies to:
\[ \frac{x + 8}{2} = \sqrt{3x + 16}. \]
Next, we square both sides again:
\[ \left(\frac{x + 8}{2}\right)^2 = 3x + 16. \]
Expanding the left side:
\[ \frac{(x + 8)^2}{4} = 3x + 16. \]
Multiplying both sides by \(4\) to eliminate the fraction:
\[ (x + 8)^2 = 12x + 64. \]
Expanding the left side gives:
\[ x^2 + 16x + 64 = 12x + 64. \]
Subtract \(12x\) and \(64\) from both sides:
\[ x^2 + 16x + 64 - 12x - 64 = 0, \]
which simplifies to:
\[ x^2 + 4x = 0. \]
Factoring out \(x\):
\[ x(x + 4) = 0. \]
Setting each factor to zero gives the solutions:
\[ x = 0 \quad \text{or} \quad x + 4 = 0 \Rightarrow x = -4. \]
We should check these solutions in the original equation to avoid extraneous solutions.
- For \(x = 0\):
\[ \sqrt{0 + 4} + \sqrt{3(0) + 16} = \sqrt{4} + \sqrt{16} = 2 + 4 = 6 \quad (\text{not a solution}). \]
- For \(x = -4\):
\[ \sqrt{-4 + 4} + \sqrt{3(-4) + 16} = \sqrt{0} + \sqrt{-12 + 16} = 0 + \sqrt{4} = 0 + 2 = 2 \quad (\text{is a solution}). \]
Thus, the only solution to the original equation is:
\[ \boxed{-4}. \]