n^2+3n-16=0

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



n^2+3n-16=0

a = 1; b = 3; c = -16;
Δ = b2-4ac
Δ = 32-4·1·(-16)
Δ = 73
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{-(3)-\sqrt{73}}{2*1}=\frac{-3-\sqrt{73}}{2}
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{73}}{2*1}=\frac{-3+\sqrt{73}}{2}
$

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