Question

z² -4_z+29=0

Answer 1

By the quadratic formula;

$z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where;

a=1

b=4

c=29

$z=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(29)}}{2(1)}$

$z=\frac{4\pm \sqrt{16-116}}{2}$

$z=\frac{4\pm \sqrt{-100}}{2}$

$z=\frac{4\pm 10i}{2}$

$z=\frac{4}{2}\pm\frac{10i}{2}$

$z=4\pm5i$

hence

$z=4+5i,\space 4-5i$