-40+2n=4n(n+8)

Solution for -40+2n=4n(n+8) equation:

-40+2n=4n(n+8)

We move all terms to the left:

-40+2n-(4n(n+8))=0


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

4n(n+8)

We multiply parentheses

4n^2+32n

Back to the equation:

-(4n^2+32n)


We get rid of parentheses

-4n^2+2n-32n-40=0

We add all the numbers together, and all the variables

-4n^2-30n-40=0

a = -4; b = -30; c = -40;
Δ = b2-4ac
Δ = -302-4·(-4)·(-40)
Δ = 260
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:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-2\sqrt{65}}{2*-4}=\frac{30-2\sqrt{65}}{-8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+2\sqrt{65}}{2*-4}=\frac{30+2\sqrt{65}}{-8}
$