Mathematics
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OpenStudy (gabylovesyou):
What are the solutions of x2 + 5x − 24 = 0 @HawkCrimson
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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
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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
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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?
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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