1/2n-4(2+n)=20

Solution for 1/2n-4(2+n)=20 equation:

1/2n-4(2+n)=20

We move all terms to the left:

1/2n-4(2+n)-(20)=0


Domain of the equation: 2n!=0

n!=0/2

n!=0

n∈R


We add all the numbers together, and all the variables

1/2n-4(n+2)-20=0

We multiply parentheses

1/2n-4n-8-20=0

We multiply all the terms by the denominator


-4n*2n-8*2n-20*2n+1=0

Wy multiply elements

-8n^2-16n-40n+1=0

We add all the numbers together, and all the variables

-8n^2-56n+1=0

a = -8; b = -56; c = +1;
Δ = b2-4ac
Δ = -562-4·(-8)·1
Δ = 3168
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:

$\sqrt{\Delta}=\sqrt{3168}=\sqrt{144*22}=\sqrt{144}*\sqrt{22}=12\sqrt{22}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-56)-12\sqrt{22}}{2*-8}=\frac{56-12\sqrt{22}}{-16}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-56)+12\sqrt{22}}{2*-8}=\frac{56+12\sqrt{22}}{-16}
$