n2+4n=32

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Solution for n2+4n=32 equation:



n2+4n=32
We move all terms to the left:
n2+4n-(32)=0
We add all the numbers together, and all the variables
n^2+4n-32=0
a = 1; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·1·(-32)
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*1}=\frac{-16}{2} =-8 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*1}=\frac{8}{2} =4 $

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