(-n-2)-(2n+5)n=0

Solution for (-n-2)-(2n+5)n=0 equation:

(-n-2)-(2n+5)n=0

We add all the numbers together, and all the variables

(-1n-2)-(2n+5)n=0

We multiply parentheses

-2n^2+(-1n-2)-5n=0

We get rid of parentheses

-2n^2-1n-5n-2=0

We add all the numbers together, and all the variables

-2n^2-6n-2=0

a = -2; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·(-2)·(-2)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{5}}{2*-2}=\frac{6-2\sqrt{5}}{-4}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{5}}{2*-2}=\frac{6+2\sqrt{5}}{-4}
$