Question

Can someone please help me answer this question (it's attached)

Answer 1

You can use this formula for any quadratic eqn: az^2 + bz + c = 0

x = ( -b +/- sqrt(b^2 - 4ac) ) /2a

So for your z = x +/- iy

x = -4/2 = -2

and

iy = sqrt(4^2 - 4*1*6)/2 = sqrt(16-24)/2 = sqrt(-8)/2 = i*sqrt(2)

z = -2 +/- i sqrt(2)

Answer 2

We know that the real number are contained in complexe set , that mean we can solve the equation in real number and we found z=(-4(+/-)squar(10))/2; with Im(z)=0.
The solotion of sreedevb is also correct ,but it's not the only solution .
i hope that you understand ,bye

Answer 3

Thank you!!