I have an equation... $6+2x^2-3x=8x^2$ I can turn it into a quadratic form like this $6x^2+3x-6$ or $-6x^2-3x+6$ Depending if I move the variable from the left to right or right to left.

Given the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The first equation is, $a = 6, b = 3, c = -b$ The second gives $a = -6, b = -3, c = 6$

But if I plug both of those into the quadratic formula I get two different answers (which makes sense). But how could we have two different answers to equivalent equations? Which one is correct? I'm sure I'm doing something totally wrong. I've been trying to figure it out for a while now.


The solutions of $-6x^2-3x+6 =0 $ are;

$x= \dfrac{3\pm\sqrt{9+144}}{-12} = \dfrac{3\pm\sqrt{153}}{-12} = \dfrac{-3+\sqrt{153}}{12}$ or $\dfrac{-3-\sqrt{153}}{12}$

While the solution of $6x^2+3x-6=0$ are;

$x= \dfrac{-3\pm\sqrt{9+144}}{12} = \dfrac{-3\pm\sqrt{153}}{12}= \dfrac{-3-\sqrt{153}}{12}$ or $\dfrac{-3+\sqrt{153}}{12}$

Which are both equal .It was just a matter of you expanding them.

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  • $\begingroup$ Ergh I knew I was doing something completely stupid. Thank you! $\endgroup$ – Coreylh Jun 7 '18 at 21:06
  • $\begingroup$ @Coreylh its alright , ive made stupider mistakes XD $\endgroup$ – The Integrator Jun 7 '18 at 21:06

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