3n^2+16n=0

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Solution for 3n^2+16n=0 equation: 



3n^2+16n=0

a = 3; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·3·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*3}=\frac{-32}{6}
=-5+1/3
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*3}=\frac{0}{6}
=0
$

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