$f(x)=x^{2}+3 x-10$

$=x^{2}+5 x-2 x-10$

$=x(x+5)-2(x+5)$

$=(x-2)(x+5)$

$\therefore \mathrm{f}(\mathrm{x})=0 \Rightarrow(\mathrm{x}-2)(\mathrm{x}+5)=0$

$\Rightarrow x-2=0$ or $x+5=0$

$\Rightarrow x=2$ or $x=-5$

So, the zeroes of $f(x)$ are 2 and $-5$. Sum of zeroes $=2+(-5)=-3=\frac{-3}{1}=\frac{-(\text { coefficient of } x)}{\left(\text { coefficient of } x^{2}\right)}$

Product of zeroes $=2 \times(-5)=-10=\frac{-10}{1}=\frac{\text { constant term }}{\left(\text { coef ficient of } x^{2}\right)}$