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Uhm it's either 6x-7x=12 which be 6x-7x=-1x then divide -1 by -1x and 12 and you should get x=-12 or it would be 9x-7x which equals 2x then divide 2 by 2x and 12 and you should get x=6
Hold on let me check
Okay so it's the second one. It's 9x-7x=12 so 9x-7x=2x then 2x is divided by 2 cross the twos out and divide 2 by 12 and you should get x= 6
Looks like you might need to use the quadratic formula: \[\Large x= \frac{-b \pm \sqrt{b^2-4ac~}}{2a} \] where a, b, and c are constants associated with your polynomial \(\large 3x^2-7x-12=0\) so a= 3, b = -7 and c = -12 \[\Large x= \frac{-(-7) \pm \sqrt{(-7)^2 - 4 (3)(-12)~}}{2(3)} \]
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@luigi0210 wrote:
Looks like you might need to use the quadratic formula:
\[\Large x= \frac{-b \pm \sqrt{b^2-4ac~}}{2a} \]
where a, b, and c are constants associated with your polynomial
\(\large 3x^2-7x-12=0\) so a= 3, b = -7 and c = -12
\[\Large x= \frac{-(-7) \pm \sqrt{(-7)^2 - 4 (3)(-12)~}}{2(3)} \]
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