Answer:NoEdit:Yes, based on original equation. (Credit to greenpumpkin for correction)Step-by-step explanation:For this problem, we simply need to find the values of x that can make the equation true. So, let's begin by isolating the "x" variable.sqrt(2x + 13) = x + 5[sqrt(2x + 13)]^2 = (x + 5)^22x + 13 = x^2 + 10x + 250 = x^2 + 8x + 12Note, we can remove the sqrt method by squaring both sides of the equation. Doing this, we see we have a quadratic equation meaning we can apply the quadratic formula to find solutions for x.[-b +/- sqrt( b^2 - 4(a)(c) ) ] / 2aLet a = 1, b = 8, and c = 12[-8 +/- sqrt( (8)^2 - 4(1)(12) ) ] / 2(1)= [-8 +/- sqrt( 64 - 48 ) ] / 2= [-8 +/- sqrt(16) ] / 2= [ -8 +/- 4 ] / 2So, x = [ -8 + 4 ] / 2 and x = [-8 - 4 ] / 2x = [-4] / 2 = -2 and x = [-12] / 2 = -6Hence, the two values of x that can solve this quadratic equation are x = -2 and x = -6.Therefore, we know that x = -6 is not extraneous, meaning it is a solution to our equation.Cheers.----------------------------------------------------Edit:Plugging the value of -6 back into the original equation, we get the following:sqrt(2x + 13) = x + 5sqrt(2(-6) + 13) = (-6) + 5sqrt (1) = -11 != -1Given that 1 cannot equal negative 1, we can say that x = -6 is an extraneous solution. (Credit to greenpumpkin for correction)