Complex number and quadratic formula?

Question

how would you solve the equation and find the roots of ix^2-3x-5i=0 ... i got (3+-sqrt9-20i^2)/2i but you cannot have i in the bottom so or maybe i did it wrong?

Answer 1

to get i off the bottom multiply the top and bottom by i

Answer 2

From ix^2-3x-5i=0

a=+i, b= -3, c= -5i

x= {-b+/-rt(b^2-4ac)} / 2a

x={3+/-rt[(9-4(i)(-5i)]} / 2i

x={3+/-rt[9+20i^2]} / 2i [Note: you got 9-20i^2 here]

x=}3+/-rt[9-20]}/ 2i

x=[3+/-rt(-11)] / 2i

x=[3+/-rt(ii)rt(-1)] /2i

x=[3+/-rt(11)i} / 2i

Now, multiply top and bottom by i

x=3i+/-rt(11)i^2 /2i^2

x=[3i-/+rt(11)] / -2

Now multiply top and bottom by -1

x=[-3i+/-rt(11)] / 2