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10n^2+11n-8=0
a = 10; b = 11; c = -8;
Δ = b2-4ac
Δ = 112-4·10·(-8)
Δ = 441
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{441}=21$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*10}=\frac{-32}{20} =-1+3/5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*10}=\frac{10}{20} =1/2 $
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