2n2+4n-16=0

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Solution for 2n2+4n-16=0 equation: 



2n^2+4n-16=0

a = 2; b = 4; c = -16;
Δ = b2-4ac
Δ = 42-4·2·(-16)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$


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


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*2}=\frac{8}{4}
=2
$

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