13n^2+4n=0

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Solution for 13n^2+4n=0 equation: 



13n^2+4n=0

a = 13; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·13·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*13}=\frac{-8}{26}
=-4/13
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*13}=\frac{0}{26}
=0
$

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