Solve the equation. \[ \sqrt{x+28}+2=x \] The solution set is

To solve the equation \( \sqrt{x+28} + 2 = x \), we first isolate the square root: \[ \sqrt{x+28} = x - 2 \] Next, we square both sides to eliminate the square root: \[ x + 28 = (x - 2)^2 \] Expanding the right side, we get: \[ x + 28 = x^2 - 4x + 4 \] Now, let's move all terms to one side of the equation: \[ 0 = x^2 - 4x - x + 4 - 28 \] This simplifies to: \[ 0 = x^2 - 5x - 24 \] Now, we can factor the quadratic equation: \[ 0 = (x - 8)(x + 3) \] Setting each factor to zero gives us: \[ x - 8 = 0 \quad \Rightarrow \quad x = 8 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] Next, we need to check the solutions in the original equation to ensure they are valid (as squaring may introduce extraneous solutions): For \( x = 8 \): \[ \sqrt{8 + 28} + 2 = \sqrt{36} + 2 = 6 + 2 = 8 \quad \text{(valid)} \] For \( x = -3 \): \[ \sqrt{-3 + 28} + 2 = \sqrt{25} + 2 = 5 + 2 = 7 \quad \text{(not valid)} \] Thus, the only solution is: \[ \boxed{8} \]