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@Vocaloid
When you factor you want the equation to be equal to 0
so this is equal
No, I mean make it so that the equation is on one side of the equation and set to 0 \(x^2-x=30\)
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You would subtract 30 from both sides
=> \(x^2-x-30=0\)
x^2-x-30?
Right
I feel comfortable with the quadratic equation, is there a specific method for solving that you would like to do?
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that is fine
\(x=\large{\frac{-b\pm \sqrt{b^2-4ac}}{2a}}\) Based on the equation we know that \(a=1\\ b=-1\\ c=-30\)
\(\large{\frac{-(-1)+\sqrt{(-1)^2-4(-30)}}{2}}\) and \(\large{\frac{-(-1)-\sqrt{(-1)^2-4(-30)}}{2}}\)
Yeah so i got b
Yeh
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Thanks!!
np :P
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