-n(n-6)-8=-(1+n)

Solution for -n(n-6)-8=-(1+n) equation:

-n(n-6)-8=-(1+n)

We move all terms to the left:

-n(n-6)-8-(-(1+n))=0

We add all the numbers together, and all the variables

-n(n-6)-(-(n+1))-8=0

We multiply parentheses

-n^2+6n-(-(n+1))-8=0


We calculate terms in parentheses: -(-(n+1)), so:

-(n+1)

We get rid of parentheses

-n-1

We add all the numbers together, and all the variables

-1n-1

Back to the equation:

-(-1n-1)


We add all the numbers together, and all the variables

-1n^2+6n-(-1n-1)-8=0

We get rid of parentheses

-1n^2+6n+1n+1-8=0

We add all the numbers together, and all the variables

-1n^2+7n-7=0

a = -1; b = 7; c = -7;
Δ = b2-4ac
Δ = 72-4·(-1)·(-7)
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{-(7)-\sqrt{21}}{2*-1}=\frac{-7-\sqrt{21}}{-2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{21}}{2*-1}=\frac{-7+\sqrt{21}}{-2}
$