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4n^2+19n-590=0
a = 4; b = 19; c = -590;
Δ = b2-4ac
Δ = 192-4·4·(-590)
Δ = 9801
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{9801}=99$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-99}{2*4}=\frac{-118}{8} =-14+3/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+99}{2*4}=\frac{80}{8} =10 $
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