16n2+n-2304=0

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Solution for 16n2+n-2304=0 equation:



16n^2+n-2304=0
a = 16; b = 1; c = -2304;
Δ = b2-4ac
Δ = 12-4·16·(-2304)
Δ = 147457
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{147457}}{2*16}=\frac{-1-\sqrt{147457}}{32} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{147457}}{2*16}=\frac{-1+\sqrt{147457}}{32} $

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