(2/3*n)+7=7+n

Solution for (2/3*n)+7=7+n equation:

(2/3n)+7=7+n

We move all terms to the left:

(2/3n)+7-(7+n)=0


Domain of the equation: 3n)!=0

n!=0/1

n!=0

n∈R


We add all the numbers together, and all the variables

(+2/3n)-(n+7)+7=0

We get rid of parentheses

2/3n-n-7+7=0

We multiply all the terms by the denominator


-n*3n-7*3n+7*3n+2=0

Wy multiply elements

-3n^2-21n+21n+2=0

We add all the numbers together, and all the variables

-3n^2+2=0

a = -3; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-3)·2
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-3}=\frac{0-2\sqrt{6}}{-6}
=-\frac{2\sqrt{6}}{-6}
=-\frac{\sqrt{6}}{-3}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-3}=\frac{0+2\sqrt{6}}{-6}
=\frac{2\sqrt{6}}{-6}
=\frac{\sqrt{6}}{-3}
$