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4n^2-5n-20=0
a = 4; b = -5; c = -20;
Δ = b2-4ac
Δ = -52-4·4·(-20)
Δ = 345
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{345}}{2*4}=\frac{5-\sqrt{345}}{8} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{345}}{2*4}=\frac{5+\sqrt{345}}{8} $
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