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6n^2+17n-3=0
a = 6; b = 17; c = -3;
Δ = b2-4ac
Δ = 172-4·6·(-3)
Δ = 361
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{361}=19$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-19}{2*6}=\frac{-36}{12} =-3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+19}{2*6}=\frac{2}{12} =1/6 $
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