n=((2/3)(n+5))

Solution for n=((2/3)(n+5)) equation:

n=((2/3)(n+5))

We move all terms to the left:

n-(((2/3)(n+5)))=0


Domain of the equation: 3)(n+5)))!=0

n∈R


We add all the numbers together, and all the variables

n-(((+2/3)(n+5)))=0

We multiply parentheses ..

-(((+2n^2+2/3*5)))+n=0

We multiply all the terms by the denominator


-(((+2n^2+2+n*3*5)))=0


We calculate terms in parentheses: -(((+2n^2+2+n*3*5))), so:

((+2n^2+2+n*3*5))


We calculate terms in parentheses: +((+2n^2+2+n*3*5)), so:

(+2n^2+2+n*3*5)

We get rid of parentheses

2n^2+n*3*5+2

Wy multiply elements

2n^2+15n*5+2

Wy multiply elements

2n^2+75n+2

Back to the equation:

+(2n^2+75n+2)


We get rid of parentheses

2n^2+75n+2

Back to the equation:

-(2n^2+75n+2)


We get rid of parentheses

-2n^2-75n-2=0

a = -2; b = -75; c = -2;
Δ = b2-4ac
Δ = -752-4·(-2)·(-2)
Δ = 5609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-75)-\sqrt{5609}}{2*-2}=\frac{75-\sqrt{5609}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-75)+\sqrt{5609}}{2*-2}=\frac{75+\sqrt{5609}}{-4}
$