n-(2n-1)=5-2n(n-1)

Solution for n-(2n-1)=5-2n(n-1) equation:

n-(2n-1)=5-2n(n-1)

We move all terms to the left:

n-(2n-1)-(5-2n(n-1))=0

We get rid of parentheses

n-2n-(5-2n(n-1))+1=0


We calculate terms in parentheses: -(5-2n(n-1)), so:

5-2n(n-1)

determiningTheFunctionDomain
-2n(n-1)+5

We multiply parentheses

-2n^2+2n+5

Back to the equation:

-(-2n^2+2n+5)


We add all the numbers together, and all the variables

-(-2n^2+2n+5)-1n+1=0

We get rid of parentheses

2n^2-2n-1n-5+1=0

We add all the numbers together, and all the variables

2n^2-3n-4=0

a = 2; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·2·(-4)
Δ = 41
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{-(-3)-\sqrt{41}}{2*2}=\frac{3-\sqrt{41}}{4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{41}}{2*2}=\frac{3+\sqrt{41}}{4}
$