Let $f\left(x\right)=a{x}^2+bx+c$

Given, $f\left(-2\right)+f\left(3\right)=0$

$\Rightarrow 4a-2b+c+9a+3b+c=0$

$\Rightarrow 13a+b+2c=0 ...\left(\mathrm{i}\right)$

Since $-1$ is a root of $f\left(x\right)=0$. So, $f\left(-1\right)=0$

$\Rightarrow a-b+c=0 ...\left(\mathrm{ii}\right)$

On solving equations $\left(\mathrm{i}\right)$ & $\left(\mathrm{ii}\right)$, we get

$11a+3b=0$  $\Rightarrow -\frac{b}{a}=\frac{11}{3}$

Now, we know that sum of the roots of $a{x}^2+bx+c=0$ is $\frac{-b}{a}$.

$\therefore$ The sum of the roots of $f\left(x\right)=0$ is $\frac{11}{3}$.