(n-3)/(2n-2)=-n

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

(n-3)/(2n-2)=-n

We move all terms to the left:

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


Domain of the equation: (2n-2)!=0

We move all terms containing n to the left, all other terms to the right

2n!=2

n!=2/2

n!=1

n∈R


We add all the numbers together, and all the variables

(n-3)/(2n-2)-(-1n)=0

We get rid of parentheses

(n-3)/(2n-2)+1n=0

We multiply all the terms by the denominator


(n-3)+1n*(2n-2)=0

We multiply parentheses

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2n^2-1n-3=0

a = 2; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·2·(-3)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*2}=\frac{-4}{4}
=-1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*2}=\frac{6}{4}
=1+1/2
$