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Mathematics 12 Online
OpenStudy (anonymous):

HOW WOULD I SOLVE; r^+2r-33=0?

OpenStudy (anonymous):

Is the equation supposed to be \[r ^{2}+2r-33=0\]

OpenStudy (anonymous):

If so, I don't think there are any any simple factors for the equation so you should use the quadratic equation \[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

OpenStudy (anonymous):

\[a=1,b=2,c=-33\] a,b,c corresponds to the coefficents of each term within the form \[ax ^{2}+bx+c=0\]

OpenStudy (solomonzelman):

\[r^2+2r-33=0\]\[r^2+2r=33\]\[r^2+2r+1=33+1\]\[(r+1)^2=34\]take it from here.

OpenStudy (anonymous):

yeah I meant r squared.

OpenStudy (anonymous):

thanks @VeritasVosLiberabit I got it.

OpenStudy (anonymous):

you're welcome

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