Answer: [tex](x-8)(x+3)=0[/tex]
Step-by-step explanation:
The standard quadratic equation is [tex]ax^2+bx+c=0[/tex]
Given quadratic equation:-[tex]x^2-5x-24=0[/tex], here a=1, b= -5, c= -24
Now, factorize given functions, first find two numbers that multiply to give ac and add to give b
ac=1×-24= -24
After checking factors of -24 we found -8 and 3 which multiply to give -24 and add to give -5
Rewrite the equation by writing -5 as sum of -8 and 3, we get
[tex]x^2+(-8+5)x-24=0\\\Rightarrow\ x^2-8x+3x-24=0\\\Rightarrow\ (x^2-8x)+(3x-24)=0\\\Rightarrow\ x(x-8)+3(x-8)=0\\\\[\text{Take (x-8) as common outside}]\\\Rightarrow\ (x-8)(x+3)=0[/tex]
