(n-5)(n-5)=n

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

(n-5)(n-5)=n

We move all terms to the left:

(n-5)(n-5)-(n)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+n^2-5n-5n+25)-1n=0

We get rid of parentheses

n^2-5n-5n-1n+25=0

We add all the numbers together, and all the variables

n^2-11n+25=0

a = 1; b = -11; c = +25;
Δ = b2-4ac
Δ = -112-4·1·25
Δ = 21
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{-(-11)-\sqrt{21}}{2*1}=\frac{11-\sqrt{21}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{21}}{2*1}=\frac{11+\sqrt{21}}{2}
$