0=(n-2)(n-6)-10(n+4)

Solution for 0=(n-2)(n-6)-10(n+4) equation:

0=(n-2)(n-6)-10(n+4)

We move all terms to the left:

0-((n-2)(n-6)-10(n+4))=0

We add all the numbers together, and all the variables

-((n-2)(n-6)-10(n+4))=0

We multiply parentheses ..

-((+n^2-6n-2n+12)-10(n+4))=0


We calculate terms in parentheses: -((+n^2-6n-2n+12)-10(n+4)), so:

(+n^2-6n-2n+12)-10(n+4)

We multiply parentheses

(+n^2-6n-2n+12)-10n-40

We get rid of parentheses

n^2-6n-2n-10n+12-40

We add all the numbers together, and all the variables

n^2-18n-28

Back to the equation:

-(n^2-18n-28)


We get rid of parentheses

-n^2+18n+28=0

We add all the numbers together, and all the variables

-1n^2+18n+28=0

a = -1; b = 18; c = +28;
Δ = b2-4ac
Δ = 182-4·(-1)·28
Δ = 436
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:

$\sqrt{\Delta}=\sqrt{436}=\sqrt{4*109}=\sqrt{4}*\sqrt{109}=2\sqrt{109}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{109}}{2*-1}=\frac{-18-2\sqrt{109}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{109}}{2*-1}=\frac{-18+2\sqrt{109}}{-2}
$