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Mathematics 8 Online
OpenStudy (anonymous):

Fan and Medal Im confused please help Solve: 5x2 + 25x = 0.

OpenStudy (anonymous):

How would I get this into quadratic formula without a c variable

OpenStudy (anonymous):

@jim_thompson5910 @Vocaloid @Hero

OpenStudy (anonymous):

Then you do not use c at all The 4ac becomes 0 in the equation.

OpenStudy (anonymous):

out of my answer choices what would it be then? I'm confused at how you would plug it in and still get the answer

OpenStudy (anonymous):

x = 0, x = 5 x = 0, x = −5 x = −5, x = 0, x = 5 x = 5, x = 25

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ax^2+bx+c so a=5 and b=25

OpenStudy (anonymous):

so it would be -25 =+/- sqrt of 25 all over 10

OpenStudy (anonymous):

i got it thanks

OpenStudy (anonymous):

\[\frac{ -25\pm \sqrt{25^2-4(5)(0)} }{ 2(5) }\] \[\frac{ -25\pm \sqrt{625} }{ 10 }\] \[\frac{ -25\pm25 }{ 10 }\] using addition it is \[\frac{ -25+25 }{ 10 }=0\] using subtraction it is \[\frac{ -25-25 }{ 10}=-5\]

OpenStudy (anonymous):

Glad you did :)

OpenStudy (anonymous):

wait it is not square root of 25. You mean square root of 625 since it is b^2 not b

OpenStudy (jack1):

all good now?

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