3n^2+n-4=0

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Solution for 3n^2+n-4=0 equation: 



3n^2+n-4=0

a = 3; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·3·(-4)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*3}=\frac{-8}{6}
=-1+1/3
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*3}=\frac{6}{6}
=1
$

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