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(3x^2)-243(x^-2)=0
We multiply parentheses
3x^2-243x+486=0
a = 3; b = -243; c = +486;
Δ = b2-4ac
Δ = -2432-4·3·486
Δ = 53217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{53217}=\sqrt{729*73}=\sqrt{729}*\sqrt{73}=27\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-243)-27\sqrt{73}}{2*3}=\frac{243-27\sqrt{73}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-243)+27\sqrt{73}}{2*3}=\frac{243+27\sqrt{73}}{6} $
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