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

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Konrad509idn
[tex]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[/tex]
Lilithidn
[tex]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}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{5-\sqrt{121}}{2 }=\frac{ 5-11}{2}=\frac{-6}{2}=-3[/tex]

[tex]x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{5+\sqrt{121}}{2 }=\frac{ 5+11}{2}=\frac{16}{2}= 8\\ \\Answer: x=-3 \ \ \ or \ \ \ x=8[/tex]





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