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6x^2-3x-1=0
a = 6; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·6·(-1)
Δ = 33
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{-(-3)-\sqrt{33}}{2*6}=\frac{3-\sqrt{33}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{33}}{2*6}=\frac{3+\sqrt{33}}{12} $
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