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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(-1.n(2n+9)), so:

-1.n(2n+9)

We multiply parentheses

-2n^2-9n

Back to the equation:

-(-2n^2-9n)


We get rid of parentheses

2n^2+9n+n=0

We add all the numbers together, and all the variables

2n^2+10n=0

a = 2; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·2·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*2}=\frac{-20}{4}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*2}=\frac{0}{4}
=0
$