12=(2/3)(4-8n)

Solution for 12=(2/3)(4-8n) equation:

12=(2/3)(4-8n)

We move all terms to the left:

12-((2/3)(4-8n))=0


Domain of the equation: 3)(4-8n))!=0

n∈R


We add all the numbers together, and all the variables

-((+2/3)(-8n+4))+12=0

We multiply parentheses ..

-((-16n^2+2/3*4))+12=0

We multiply all the terms by the denominator


-((-16n^2+2+12*3*4))=0


We calculate terms in parentheses: -((-16n^2+2+12*3*4)), so:

(-16n^2+2+12*3*4)

We get rid of parentheses

-16n^2+2+12*3*4

We add all the numbers together, and all the variables

-16n^2+146

Back to the equation:

-(-16n^2+146)


We get rid of parentheses

16n^2-146=0

a = 16; b = 0; c = -146;
Δ = b2-4ac
Δ = 02-4·16·(-146)
Δ = 9344
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:

$\sqrt{\Delta}=\sqrt{9344}=\sqrt{64*146}=\sqrt{64}*\sqrt{146}=8\sqrt{146}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{146}}{2*16}=\frac{0-8\sqrt{146}}{32}
=-\frac{8\sqrt{146}}{32}
=-\frac{\sqrt{146}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{146}}{2*16}=\frac{0+8\sqrt{146}}{32}
=\frac{8\sqrt{146}}{32}
=\frac{\sqrt{146}}{4}
$