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3x^2-13x-3=0
a = 3; b = -13; c = -3;
Δ = b2-4ac
Δ = -132-4·3·(-3)
Δ = 205
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{-(-13)-\sqrt{205}}{2*3}=\frac{13-\sqrt{205}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+\sqrt{205}}{2*3}=\frac{13+\sqrt{205}}{6} $
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