Question 435098
2-5i to be root of equation {{{z^2-4z+29=0}}}, must satisfy this equation.

Plug z=2-5i, {{{(2-5i)^2-4(2-5i)+29=0}}}, transform the left side. 

{{{4-20i-25-8+20i+29=0}}}, combine like terms.

{{{(-20+20)i-4-25+29=0}}},

{{{0i-29+29=0}}}

{{{0=0}}}

Answer:The value z=2-5i satisfy our equation, therefore is its root.