Let f(x)=ax2+bx+c
Given, f(−2)+f(3)=0
⇒4a−2b+c+9a+3b+c=0
⇒13a+b+2c=0...(i)
Since −1 is a root of f(x)=0. So, f(−1)=0
⇒a−b+c=0...(ii)
On solving equations (i) & (ii), we get
11a+3b=0 ⇒−ab=311
Now, we know that sum of the roots of ax2+bx+c=0 is a−b.
∴ The sum of the roots of f(x)=0 is 311.