$$(m+1)x^2+2(m-1)x+4m+1=0$$ 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?
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You are trying to find which values of $m$ satisfy $8m^2 + 28m > 0$. Have you tried completing the square? – fractal1729 Dec 21 '17 at 20:46
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@fractal1729 What do you mean by completing the square ? – RiktasMath Dec 21 '17 at 20:47
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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 $-12m^2-28m$ ...
Michael Hoppe
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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)
Higurashi
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