0=n^2-4n

Solution for 0=n^2-4n equation:

0=n^2-4n

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(n^2-4n)=0

We get rid of parentheses

-n^2+4n=0

We add all the numbers together, and all the variables

-1n^2+4n=0

a = -1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-1)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-1}=\frac{-8}{-2}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-1}=\frac{0}{-2}
=0
$