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5n^2+21n-20=0
a = 5; b = 21; c = -20;
Δ = b2-4ac
Δ = 212-4·5·(-20)
Δ = 841
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{841}=29$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-29}{2*5}=\frac{-50}{10} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+29}{2*5}=\frac{8}{10} =4/5 $
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