n^2+4n-192=0

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Solution for n^2+4n-192=0 equation:



n^2+4n-192=0
a = 1; b = 4; c = -192;
Δ = b2-4ac
Δ = 42-4·1·(-192)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-28}{2*1}=\frac{-32}{2} =-16 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+28}{2*1}=\frac{24}{2} =12 $

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