Find the zeros of the quadratic polynomial

We have:

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

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

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

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

$\therefore f(x)=0=>(x-2)(x+5)=0$

$=>x-2=0$ or $x+5=0$

$=>x=2$ or $x=-5$

So, the zeroes of $f(x)$ are 2 and $-5$.

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

Product of the zeroes $=2 \times(-5)=-10=\frac{-10}{1}=\frac{\text { constant term }}{\text { (coefficient of } x^{2} \text { ) }}$