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3x^2-196=217x
We move all terms to the left:
3x^2-196-(217x)=0
a = 3; b = -217; c = -196;
Δ = b2-4ac
Δ = -2172-4·3·(-196)
Δ = 49441
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{49441}=\sqrt{49*1009}=\sqrt{49}*\sqrt{1009}=7\sqrt{1009}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-217)-7\sqrt{1009}}{2*3}=\frac{217-7\sqrt{1009}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-217)+7\sqrt{1009}}{2*3}=\frac{217+7\sqrt{1009}}{6} $
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