Let $a > 0, b > 0$ and $c > 0$. Then both the roots of the equation

$2a{x}^2+3bx+5c=0$                    (1)

The discriminant $D$ of the equation (1) is given by

$D=(3b{)}^2-4(2a\left)\right(5c)=9b^2-40ac$

If $D \ge 0$, then both the roots of (1) are real.

Also, since $ac>0,D<9b^2$

$\Rightarrow \sqrt{D}<3b[\because b>0]$

Therefore, in this case both the roots  $\frac{-3b-\sqrt{D}}{4a}$ and

$\frac{-3b+\sqrt{D}}{4a}$ are negative.

If $D < 0$, both the roots of (1) are imaginary and are given by

$x=\frac{-3b\pm i\sqrt{40ac-9b^2}}{2}$

Both these roots have negative real parts.