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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+8n^2-32n+4n-16)=0

We get rid of parentheses

8n^2-32n+4n-16=0

We add all the numbers together, and all the variables

8n^2-28n-16=0

a = 8; b = -28; c = -16;
Δ = b2-4ac
Δ = -282-4·8·(-16)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-36}{2*8}=\frac{-8}{16}
=-1/2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+36}{2*8}=\frac{64}{16}
=4
$