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OpenStudy (anonymous):

solve by factoring 2x^2+3x=0

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OpenStudy (a_clan):

x(2x+3)=0 x=0,-3/2

OpenStudy (anonymous):

there's supposed to be 2( )

OpenStudy (anonymous):

2x^2 + 3x take x as common x(2x+3)=0 either x=0 or 2x+3=0 now simplify 2x+3=0 transpose 3 to RHS u would get: 2x=-3 divide both sides by 2 u would get: 2x/2=-3/2 x=-3/2 hence it has two solutions these are: x=0 and x=-3/2

OpenStudy (anonymous):

understood tiny dancer or any problem

OpenStudy (anonymous):

its supposed to be shaped like this: ( ) ( )

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OpenStudy (annon):

x(2x+3)=0 2x+3=0 x=-3/2

OpenStudy (anonymous):

TinyDancer - I understand what you're asking. In this case, you are factoring out the x because there are only 2 terms, not 3. When there are 2 terms the ( ) ( ) format will not work.

OpenStudy (anonymous):

oh okay

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