47n2+5n=0

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Solution for 47n2+5n=0 equation: 



47n^2+5n=0

a = 47; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·47·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*47}=\frac{-10}{94}
=-5/47
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*47}=\frac{0}{94}
=0
$

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