(n^2)-4(n)+5=n

Solution for (n^2)-4(n)+5=n equation:

(n^2)-4(n)+5=n

We move all terms to the left:

(n^2)-4(n)+5-(n)=0

We add all the numbers together, and all the variables

n^2-5n+5=0

a = 1; b = -5; c = +5;
Δ = b2-4ac
Δ = -52-4·1·5
Δ = 5
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{-(-5)-\sqrt{5}}{2*1}=\frac{5-\sqrt{5}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{5}}{2*1}=\frac{5+\sqrt{5}}{2}
$