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x^2/7 +x = -8/7 need to use quadratic formula :/
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x^2 + 7x = -8 x^2 + 7x + 8 = 0
the x^2 is the numerator
\[\frac{1}{7}x^2 +x +8/7= 0\] \[a = \frac{1}{7}, b = 1, c = \frac{8}{7}\] In fact, get rid of those fractions \[7\cdot(\frac{1}{7}x^2 +x +8/7)= 7\cdot0 \Rightarrow x^2 + 7x + 8 = 0\] Now \[a = 1 , b = 7 , c = 8 \] \[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[\frac{-7 \pm \sqrt{49-4(8)}}{2}\] x = \[\frac{-7 \pm \sqrt{17}}{2}\]
How do i get rid of the fractions? thanks :)
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