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 :
![y=x^(2)-5x-24](https://img.qammunity.org/2019/formulas/mathematics/high-school/m9jvqfuk9qwkvam25ysvcc5wa1m1oay1jb.png)
Comparing it with quadratic equation
![y=ax^(2)+bx+c](https://img.qammunity.org/2019/formulas/mathematics/high-school/k7oxldzlkxbmmoe7zqfzb1unuxfzzz7vqu.png)
We have a=1,b=-5,c=-24.
To find the zeros we substitute y=0.
![x^(2)-5x-24=0.](https://img.qammunity.org/2019/formulas/mathematics/high-school/bcqzs7zlk4x291ph2n5m3sap86j0n4bb6z.png)
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.