Solve $LaTeX: \displaystyle x + 2 = \sqrt{2 x + 12}$ .

Squaring both sides gives $LaTeX: \displaystyle x^{2} + 4 x + 4 = 2 x + 12$ . The equation is quadratic setting it equal to zero gives $LaTeX: \displaystyle x^{2} + 2 x - 8 = 0$ . Factoring gives $LaTeX: \displaystyle (x - 2)(x + 4)=0$ so the possible solutions are $LaTeX: \displaystyle x = 2$ and $LaTeX: \displaystyle x = -4$ . Checking the solution $LaTeX: \displaystyle x = 2$ in the original equation gives $LaTeX: \displaystyle 4 = 4$ . The solution checks, so $LaTeX: \displaystyle x = 2$ is a true solution. Checking the solution $LaTeX: \displaystyle x = -4$ in the original equation gives $LaTeX: \displaystyle -2 = 2$ . The solution does no check, so $LaTeX: \displaystyle x = -4$ is an extraneous solution.