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6n^2+2n-20=0
a = 6; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·6·(-20)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*6}=\frac{-24}{12} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*6}=\frac{20}{12} =1+2/3 $
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