1n(-2n-9)=(-n-2)

Solution for 1n(-2n-9)=(-n-2) equation:

1n(-2n-9)=(-n-2)

We move all terms to the left:

1n(-2n-9)-((-n-2))=0

We add all the numbers together, and all the variables

1n(-2n-9)-((-1n-2))=0

We multiply parentheses

-2n^2-9n-((-1n-2))=0


We calculate terms in parentheses: -((-1n-2)), so:

(-1n-2)

We get rid of parentheses

-1n-2

Back to the equation:

-(-1n-2)


We get rid of parentheses

-2n^2-9n+1n+2=0

We add all the numbers together, and all the variables

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

a = -2; b = -8; c = +2;
Δ = b2-4ac
Δ = -82-4·(-2)·2
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{5}}{2*-2}=\frac{8-4\sqrt{5}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{5}}{2*-2}=\frac{8+4\sqrt{5}}{-4}
$