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6x^2-8x-4=0
a = 6; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·6·(-4)
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{10}}{2*6}=\frac{8-4\sqrt{10}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{10}}{2*6}=\frac{8+4\sqrt{10}}{12} $
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