Question 396457
Letting {{{u = x^2}}},


{{{u^2 + 14u - 32 = 0}}}


{{{(u+16)(u-2) = 0}}}


u = -16 or u = 2


Taking positive and negative square roots of both values for u, we get


x = {{{4i}}}, {{{-4i}}}, {{{sqrt(2)}}}, {{{-sqrt(2)}}}.


The imaginary roots are 4i and -4i.