Solution of ax^2+bx+c=0 is given by a quadratic formula as x=(-b+-sqrt(b^2-4ac))/(2a)
therefore for x^2+ix-i=0
x=(-i+-sqrt(i^2+4i))/2
= (-i+-sqrt(-1+4i))/2
As -1+4i=sqrt17(costheta+isintheta), where tantheta=-4
and using DeMoivre's theorem
sqrt(-1+4i)=17^(1/4)(cosalpha+isinalpha), where theta=2alpha
As tan2alpha= (2tanalpha)/(1-tan^2alpha)=-4
we have 4tan^2alpha-2tanalpha-4=0 and
tanalpha=(2+-2sqrt17)/4 and alpha=tan^(-1)((2+-2sqrt17)/4)
Hence we have two roots of -1+4i given by
17^(1/4)(cosalpha+isinalpha), where alpha=tan^(-1)((2+-2sqrt17)/4)
and x=(-i+-17^(1/4)(cosalpha+isinalpha))/2
= 17^(1/4)/2cosalpha+i(17^(1/4)/2sinalpha-1)
and 17^(1/4)/2cosalpha-i(17^(1/4)/2sinalpha+1),
where alpha=tan^(-1)((2+-2sqrt17)/4)