8(6n-4)=4n(6n-10n)+40

Solution for 8(6n-4)=4n(6n-10n)+40 equation:

8(6n-4)=4n(6n-10n)+40

We move all terms to the left:

8(6n-4)-(4n(6n-10n)+40)=0

We add all the numbers together, and all the variables

8(6n-4)-(4n(-4n)+40)=0

We multiply parentheses

48n-(4n(-4n)+40)-32=0


We calculate terms in parentheses: -(4n(-4n)+40), so:

4n(-4n)+40

We multiply parentheses

-16n^2+40

Back to the equation:

-(-16n^2+40)


We get rid of parentheses

16n^2+48n-40-32=0

We add all the numbers together, and all the variables

16n^2+48n-72=0

a = 16; b = 48; c = -72;
Δ = b2-4ac
Δ = 482-4·16·(-72)
Δ = 6912
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:

$\sqrt{\Delta}=\sqrt{6912}=\sqrt{2304*3}=\sqrt{2304}*\sqrt{3}=48\sqrt{3}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-48\sqrt{3}}{2*16}=\frac{-48-48\sqrt{3}}{32}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+48\sqrt{3}}{2*16}=\frac{-48+48\sqrt{3}}{32}
$