8=n(n-4)

Solution for 8=n(n-4) equation:

8=n(n-4)

We move all terms to the left:

8-(n(n-4))=0


We calculate terms in parentheses: -(n(n-4)), so:

n(n-4)

We multiply parentheses

n^2-4n

Back to the equation:

-(n^2-4n)


We get rid of parentheses

-n^2+4n+8=0

We add all the numbers together, and all the variables

-1n^2+4n+8=0

a = -1; b = 4; c = +8;
Δ = b2-4ac
Δ = 42-4·(-1)·8
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3}}{2*-1}=\frac{-4-4\sqrt{3}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3}}{2*-1}=\frac{-4+4\sqrt{3}}{-2}
$