4n^2+n-4=0

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



4n^2+n-4=0

a = 4; b = 1; c = -4;
Δ = b2-4ac
Δ = 12-4·4·(-4)
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{65}}{2*4}=\frac{-1-\sqrt{65}}{8}
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*4}=\frac{-1+\sqrt{65}}{8}
$

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