Correct Answer - C
Roots of `ax^(2) + bc + c = 0` are given by `x = - (b)/(2a) +-(sqrt(b^(2)-4ac))/(2a)`
We have, `a, b gt 0 rArr -b//2a lt 0`.
If `b^(2) - 4ac lt 0`, then roots are imaginary of which real part is negative.
If `b^(2) - 4ac gt 0`, then roots are real and negative since `sqrt(b^(2) - 4ac) lt b`.
Hence, in either case, both the roots have negative real part.