# Question-76700

Question Number 76700 by Mikael_786 last updated on 29/Dec/19
Commented by benjo 1/2 santuyy last updated on 29/Dec/19
$${what}\:{are}\:{you}\:{looking}\:{for}\:,{sir}? \\$$
Commented by Mikael_786 last updated on 29/Dec/19
$${I}'{m}\:{looking}\:{for}\:{solution}\:{of}\: \\$$$${this}\:{equation}\:{Sir}. \\$$
Answered by john santu last updated on 29/Dec/19
Commented by Mikael_786 last updated on 29/Dec/19
$${thank}\:{you}\:{Sir} \\$$
Commented by MJS last updated on 29/Dec/19
$$\mathrm{there}\:\mathrm{are}\:\mathrm{more}\:\mathrm{solutions} \\$$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{y},\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{2}\right),\:\mathrm{solve}\:\mathrm{for} \\$$$${z}\:\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\$$$$\mathrm{we}\:\mathrm{end}\:\mathrm{up}\:\mathrm{with}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{16}\:\mathrm{for}\:{x} \\$$$${x}_{\mathrm{1},\:\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\$$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{39}}}{\mathrm{2}}\mathrm{i} \\$$$$\mathrm{and}\:\mathrm{6}\:\mathrm{more}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex} \\$$$$\mathrm{solutions}\:\mathrm{only}\:\mathrm{approximately} \\$$
Commented by Mikael_786 last updated on 29/Dec/19
$${thank}\:{you}\:{Sir}. \\$$