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3x^2-6x-1=0
a = 3; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·3·(-1)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{3}}{2*3}=\frac{6-4\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{3}}{2*3}=\frac{6+4\sqrt{3}}{6} $
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