n^2+19n-4070=0

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Solution for n^2+19n-4070=0 equation:



n^2+19n-4070=0
a = 1; b = 19; c = -4070;
Δ = b2-4ac
Δ = 192-4·1·(-4070)
Δ = 16641
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16641}=129$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-129}{2*1}=\frac{-148}{2} =-74 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+129}{2*1}=\frac{110}{2} =55 $

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