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Mathematics 20 Online
OpenStudy (gabylovesyou):

What are the solutions of x2 + 5x − 24 = 0 @HawkCrimson

OpenStudy (anonymous):

use the quadratic equation

OpenStudy (anonymous):

let me know if u need help

OpenStudy (anonymous):

u remember the quadratic formula?

OpenStudy (gabylovesyou):

y = ax^2 + bx + c

OpenStudy (gabylovesyou):

@HawkCrimson

OpenStudy (anonymous):

(x-3)(x+8)=0

OpenStudy (anonymous):

\[x = \frac{ -b \sqrt{b ^{2}-4ac} }{ 2a }\]

OpenStudy (anonymous):

or x=3 or x=-8

OpenStudy (gabylovesyou):

O.o @HawkCrimson

OpenStudy (anonymous):

@HawkCrimson i think or made a typo buddy

OpenStudy (gabylovesyou):

@ekmath how did u get that ?

OpenStudy (anonymous):

\[(-b+\sqrt{b^2-4ac}) / 2a , and, (-b-\sqrt{b^2-4ac}) / 2a\]

OpenStudy (gabylovesyou):

OMFG i dont know that ;(

OpenStudy (anonymous):

don't worry, we will go through it step by step

OpenStudy (anonymous):

can you tell me what A, B, and C are?

OpenStudy (anonymous):

just reduce your eqn. to x2+8x-3x-24=0 or x(x+8)-3(x+8)=0

OpenStudy (anonymous):

or (x-3)(x+8)=0

OpenStudy (anonymous):

can we only have one person speaking at a time -_- and @ekmath i knew i made that typo but was to lazy to add the plus or minus sign in

OpenStudy (anonymous):

@ekmath be my guess to explain it though :)

OpenStudy (anonymous):

@rahulon 's method totally works, but it might be hard to figure out how to apply it on a test

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