Solve $LaTeX: \displaystyle x + 2 = \sqrt{7 x + 32}$ .

Squaring both sides gives $LaTeX: \displaystyle x^{2} + 4 x + 4 = 7 x + 32$ . The equation is quadratic setting it equal to zero gives $LaTeX: \displaystyle x^{2} - 3 x - 28 = 0$ . Factoring gives $LaTeX: \displaystyle (x - 7)(x + 4)=0$ so the possible solutions are $LaTeX: \displaystyle x = 7$ and $LaTeX: \displaystyle x = -4$ . Checking the solution $LaTeX: \displaystyle x = 7$ in the original equation gives $LaTeX: \displaystyle 9 = 9$ . The solution checks, so $LaTeX: \displaystyle x = 7$ 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.