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20n^2+n-12=0
a = 20; b = 1; c = -12;
Δ = b2-4ac
Δ = 12-4·20·(-12)
Δ = 961
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{961}=31$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*20}=\frac{-32}{40} =-4/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*20}=\frac{30}{40} =3/4 $
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