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

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}
$