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2n^2+n-21=0
a = 2; b = 1; c = -21;
Δ = b2-4ac
Δ = 12-4·2·(-21)
Δ = 169
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{169}=13$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*2}=\frac{-14}{4} =-3+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*2}=\frac{12}{4} =3 $
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