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3x^2-2(3)x+(3)=0
a = 3; b = -23; c = +3;
Δ = b2-4ac
Δ = -232-4·3·3
Δ = 493
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{-(-23)-\sqrt{493}}{2*3}=\frac{23-\sqrt{493}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{493}}{2*3}=\frac{23+\sqrt{493}}{6} $
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