16-2n=-8+4n(n-3)

Solution for 16-2n=-8+4n(n-3) equation:

16-2n=-8+4n(n-3)

We move all terms to the left:

16-2n-(-8+4n(n-3))=0


We calculate terms in parentheses: -(-8+4n(n-3)), so:

-8+4n(n-3)

determiningTheFunctionDomain
4n(n-3)-8

We multiply parentheses

4n^2-12n-8

Back to the equation:

-(4n^2-12n-8)


We get rid of parentheses

-4n^2-2n+12n+8+16=0

We add all the numbers together, and all the variables

-4n^2+10n+24=0

a = -4; b = 10; c = +24;
Δ = b2-4ac
Δ = 102-4·(-4)·24
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*-4}=\frac{-32}{-8}
=+4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*-4}=\frac{12}{-8}
=-1+1/2
$