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3x^2-27x+50=0
a = 3; b = -27; c = +50;
Δ = b2-4ac
Δ = -272-4·3·50
Δ = 129
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{129}}{2*3}=\frac{27-\sqrt{129}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+\sqrt{129}}{2*3}=\frac{27+\sqrt{129}}{6} $
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