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6x^2-27x+10=0
a = 6; b = -27; c = +10;
Δ = b2-4ac
Δ = -272-4·6·10
Δ = 489
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-\sqrt{489}}{2*6}=\frac{27-\sqrt{489}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{489}}{2*6}=\frac{27+\sqrt{489}}{12} $
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