Question

How do I solve b^2-5b-24=0

Answer 1

`b^2-5b-24=0\\ b^2-8b+3b-24=0\\ b(b-8)+3(b-8)=0\\ (b+3)(b-8)=0\\ b=-3 \vee b=8`

Answer 2

`b^2-5b-24=0\\ \\a=1 , \ b=-5, \ c= -24 \\ \\ \Delta =b^2-4ac = (-5)^2 -4\cdot1\cdot (-24) = 25 +96 =121 \\ \\x_(1)=(-b-√(\Delta) )/(2a)=(5-√(121))/(2 )=( 5-11)/(2)=(-6)/(2)=-3`


`x_(2)=(-b+√(\Delta) )/(2a)=(5+√(121))/(2 )=( 5+11)/(2)=(16)/(2)= 8\\ \\Answer: x=-3 \ \ \ or \ \ \ x=8`