n(n+2)=-10(n+4)+8

Solution for n(n+2)=-10(n+4)+8 equation:

n(n+2)=-10(n+4)+8

We move all terms to the left:

n(n+2)-(-10(n+4)+8)=0

We multiply parentheses

n^2+2n-(-10(n+4)+8)=0


We calculate terms in parentheses: -(-10(n+4)+8), so:

-10(n+4)+8

We multiply parentheses

-10n-40+8

We add all the numbers together, and all the variables

-10n-32

Back to the equation:

-(-10n-32)


We get rid of parentheses

n^2+2n+10n+32=0

We add all the numbers together, and all the variables

n^2+12n+32=0

a = 1; b = 12; c = +32;
Δ = b2-4ac
Δ = 122-4·1·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4}{2*1}=\frac{-16}{2}
=-8
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4}{2*1}=\frac{-8}{2}
=-4
$