$\text{ If }a+b+c=0\text{ then the roots of the equation }4a{x}^2+3bx+2c=0\text{ where }a,b,c\in R\text{ are }$

$\text{ For given equation }4a{x}^2+3bx+2c=0$

$\begin{array}{l}\text{ we have }\mathrm{D}=(3\mathrm{b}{)}^2-4(4\mathrm{a}\left)\right(2\mathrm{c})\\ =9{\mathrm{b}}^2-32\mathrm{ac}\\ =9(-\mathrm{a}-\mathrm{c}{)}^2-32\mathrm{ac}\\ =9{\mathrm{a}}^2-14\mathrm{ac}+9{\mathrm{c}}^2\\ =9{\mathrm{c}}^2\left({\left(\frac{\mathrm{a}}{\mathrm{c}}\right)}^2-\frac{14}{9}\frac{\mathrm{a}}{\mathrm{c}}+1\right)\\ =9\left({\left(\frac{\mathrm{a}}{\mathrm{c}}-\frac{7}{9}\right)}^2-\frac{49}{81}+1\right)\end{array}$

which is always positive Hence roots are real and distinct