The roots of a quadratic equation ax^2 + bx + c = 0 are `(-b+-sqrt(b^2-4ac))/(2a)` . These are complex if b^2 - 4ac < 0

For the equation 3x^2 + bx - 5 = 0 the roots are complex if:

b^2 - 4*-5*3 < 0

=> b^2 + 60 < 0

=> b^2 < -60

But the square of a number is never negative.

**For no value of b are the roots of the equation 3x^2 + bx - 5 = 0 complex.**

3x^2 + bx - 5 = 0

The roots of the equation `3x^2+bx-5=0` are

`x=(-b+-sqrt(b^2-4xx3xx(-5)))/(2xx3)`

``Let us suppose x is complex no.

So x will complex if

`b^2+20<0`

`b^2<-20`

`b^2<20i^2`

`b<sqrt(20i^2)`

`b<isqrt(20)`

which is contradiction to fact that b is real no .Also there does not exist order relation in complex no. system . So our assumption is wrong.

So this equation has no complex root for any real value of b.