Let a > 0, b > 0  and  c > 0. Then both the roots of the equation2ax2+3bx+5c=0                    (1)

The discriminant D of the equation (1) is given byD=(3b)2−4(2a)(5c)=9b2−40acIf D ≥ 0, then both the roots of (1) are real.Also, since ac>0,D<9b2⇒ D<3b [∵b>0]Therefore, in this case both the roots  −3b−D4a and−3b+D4a are negative.If D < 0, both the roots of (1) are imaginary and are given byx=−3b±i40ac−9b22Both these roots have negative real parts.