Question 390276
Letting {{{z = w^2}}}, substitute:


{{{z^2 - 18z - 2 = 0}}}


By the quadratic formula, {{{z = (18 +- sqrt(332))/2 = 9 +- sqrt(83)}}} If {{{z = 9 + sqrt(83)}}} then {{{w = 0 +- sqrt(9 + sqrt(83))}}}. Likewise, if {{{z = 9 - sqrt(83)}}} then {{{w = 0 +- sqrt(9 - sqrt(83))}}} (this gives two real and two complex values for w).