Solve the equation. \( x=\sqrt{3 x+12}+2 \) \( x=[?] \)

To solve the equation \( x = \sqrt{3x + 12} + 2 \), start by isolating the square root: \[ x - 2 = \sqrt{3x + 12} \] Next, square both sides to eliminate the square root: \[ (x - 2)^2 = 3x + 12 \] Expanding the left side gives: \[ x^2 - 4x + 4 = 3x + 12 \] Now, rearrange the equation: \[ x^2 - 4x - 3x + 4 - 12 = 0 \] This simplifies to: \[ x^2 - 7x - 8 = 0 \] Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = -7, c = -8 \): \[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot -8}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2} \] This gives us two potential solutions: \[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-2}{2} = -1 \] Finally, we need to check both solutions in the original equation to confirm they are valid. Substituting \( x = 8 \): \[ 8 = \sqrt{3(8) + 12} + 2 \implies 8 = \sqrt{36} + 2 \implies 8 = 6 + 2 \implies 8 = 8 \quad \text{(valid)} \] Now for \( x = -1 \): \[ -1 = \sqrt{3(-1) + 12} + 2 \implies -1 = \sqrt{9} + 2 \implies -1 = 3 + 2 \implies -1 = 5 \quad \text{(not valid)} \] Thus, the solution to the equation is: \( x = 8 \)