The zeros of any equation lie on the x axis.On x axis y=0.The zeros can be found by substituting y=0 and finding the corresponding y values.
The given equation is :

Comparing it with quadratic equation

We have a=1,b=-5,c=-24.
To find the zeros we substitute y=0.

WE need to find two numbers which on multiplication will give us ac that is 1(-24)=-24.The same numbers on addition should give us b=-5.
The two numbers are -8 and 3.-8 times 3=-24 and -8+3=-5.
The factors of the equation are (x+8)(x-3)=0
making both factors =0 we have:
x+8=0 and x-3=0
Solving for x
x=-8 and x=3.
The two zeros of the equation are -3,8.