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21n^2+8n-5=0
a = 21; b = 8; c = -5;
Δ = b2-4ac
Δ = 82-4·21·(-5)
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*21}=\frac{-30}{42} =-5/7 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*21}=\frac{14}{42} =1/3 $
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