n^2+n-96=0

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Solution for n^2+n-96=0 equation: 



n^2+n-96=0

a = 1; b = 1; c = -96;
Δ = b2-4ac
Δ = 12-4·1·(-96)
Δ = 385
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{385}}{2*1}=\frac{-1-\sqrt{385}}{2}
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{385}}{2*1}=\frac{-1+\sqrt{385}}{2}
$

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