# Determining the nature of quadratic equation

If $$a,b,c$$ are real numbers and $$a+b+c =0$$ then how to prove that the equation $$4ax^2+3bx+2c$$ has two real roots. I just know that for real roots the quadratic equation should have its Discriminant greater than or equal to zero but how can is use the condition that $$a+b+c=0$$ . Any hint might help .Thanks

\begin{align}(3b)^2-4(4a)(2c)&=9b^2-32ac \\ &=9b^2+32(b+c)c\end{align}
Alternatively, the discriminant is $$9b^2-32a(-a-b)=b^2+8(2a+b)^2>0,$$ since $$a\neq 0$$ and at least one of $$b\neq 0,\; b\neq -2a$$ holds.