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Mathematics 10 Online
OpenStudy (anonymous):

x^2-6x=9? 5x^2=6+2x? x^2=4x+21? 3x^2+2x=3? I am trying to solve for x and I don't know how to solve these quadratic equations.

OpenStudy (anonymous):

ALL TOGTHER ?

OpenStudy (anonymous):

no each one is a seperate problem

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

FOR equatiom # 3 : x^2-4x-21=0 (x-7)(x+3)=0 then x=7 and x=-3

OpenStudy (anonymous):

i think x=1 on the last question but im not sure

OpenStudy (anonymous):

ok wait iam try :)

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

\[x=(1+\sqrt{10})\div 3 and x=(1-\sqrt{10})\div3\] for last one

OpenStudy (anonymous):

i solve it by The general law ..

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