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4n^2+47n=0
a = 4; b = 47; c = 0;
Δ = b2-4ac
Δ = 472-4·4·0
Δ = 2209
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{2209}=47$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-47}{2*4}=\frac{-94}{8} =-11+3/4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+47}{2*4}=\frac{0}{8} =0 $
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