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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.
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ALL TOGTHER ?
no each one is a seperate problem
OK
FOR equatiom # 3 : x^2-4x-21=0 (x-7)(x+3)=0 then x=7 and x=-3
i think x=1 on the last question but im not sure
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ok wait iam try :)
thank you
\[x=(1+\sqrt{10})\div 3 and x=(1-\sqrt{10})\div3\] for last one
i solve it by The general law ..
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