2n(n+4)=96+2(2n+20)

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Solution for 2n(n+4)=96+2(2n+20) equation:



2n(n+4)=96+2(2n+20)
We move all terms to the left:
2n(n+4)-(96+2(2n+20))=0
We multiply parentheses
2n^2+8n-(96+2(2n+20))=0
We calculate terms in parentheses: -(96+2(2n+20)), so:
96+2(2n+20)
determiningTheFunctionDomain 2(2n+20)+96
We multiply parentheses
4n+40+96
We add all the numbers together, and all the variables
4n+136
Back to the equation:
-(4n+136)
We get rid of parentheses
2n^2+8n-4n-136=0
We add all the numbers together, and all the variables
2n^2+4n-136=0
a = 2; b = 4; c = -136;
Δ = b2-4ac
Δ = 42-4·2·(-136)
Δ = 1104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{1104}=\sqrt{16*69}=\sqrt{16}*\sqrt{69}=4\sqrt{69}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{69}}{2*2}=\frac{-4-4\sqrt{69}}{4} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{69}}{2*2}=\frac{-4+4\sqrt{69}}{4} $

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