A quadratic equation $y=(k+1)x^2-3x+(k+1)$ we need to the find the set of values of $k$ for which the curve $y$ lies below the $x-$ axis.

I used the quadratic formula and equate it to $0$

$ 3\pm \frac{\sqrt{ 9-4(k+1)^2}}{2(k+1)}=0 $

PS assist, how to solve to get further.

Thanks,
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