2n=4(n-8)n=

Solution for 2n=4(n-8)n= equation:

2n=4(n-8)n=

We move all terms to the left:

2n-(4(n-8)n)=0


We calculate terms in parentheses: -(4(n-8)n), so:

4(n-8)n

We multiply parentheses

4n^2-32n

Back to the equation:

-(4n^2-32n)


We get rid of parentheses

-4n^2+2n+32n=0

We add all the numbers together, and all the variables

-4n^2+34n=0

a = -4; b = 34; c = 0;
Δ = b2-4ac
Δ = 342-4·(-4)·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-34}{2*-4}=\frac{-68}{-8}
=8+1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+34}{2*-4}=\frac{0}{-8}
=0
$