Determine parameter m so equation has complex results for x:$(m+1)x^2+2(m-1)x+4m+1=0$

Question

$$(m+1)x^2+2(m-1)x+4m+1=0$$ So for results to be complex $b^2-4ac<0$

I then get to a part where I don't know what to do next: $-8m^2-28m<0$. How do I solve m?

Answer 1

You will need $3m^2+7m>0$, so $m(m+\frac{7}{3})>0$. For this to be true, you will need to have $m$ and $(m+\frac{7}{3})$ of the same sign. Find values of $m$ that satisfy this condition.

Edit: you had a mistake in your calculations (Thanks to Michael Hoppe for seeing it)

Answer 2

Consider $-8m^2-28m$ as a parabola; its zeroes are $0$ and $-7/2$. As it is downward open it will take negative values if $m<-7/2$ or $m>0$.

BTW I've got $-12m^2-28m$ ...