q - 6 = sqrt(27 - 2q)
Squaring both sides:-
(q - 6)^2 = 27 - 2q
q^2 - 12q + 36 = 27 - 2q
q^2 - 10q + 9 = 0
(q - 9)(q - 1) = 0
so there are 2 solutions q = 1 and q = 9.
Testing to see if the solutions are not extraneous:-
q = 9:-
9 - 6 = 3
sqrt(27-18) = 3 So q = 9 is a solution (answer)
q = 1
q - 6 = -5
sqrt (27 - 2) = sqrt 25 = -5 so q = 1 is also an answer
