3n(2n-4)=36n-8n+8

Solution for 3n(2n-4)=36n-8n+8 equation:

3n(2n-4)=36n-8n+8

We move all terms to the left:

3n(2n-4)-(36n-8n+8)=0

We add all the numbers together, and all the variables

3n(2n-4)-(28n+8)=0

We multiply parentheses

6n^2-12n-(28n+8)=0

We get rid of parentheses

6n^2-12n-28n-8=0

We add all the numbers together, and all the variables

6n^2-40n-8=0

a = 6; b = -40; c = -8;
Δ = b2-4ac
Δ = -402-4·6·(-8)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-16\sqrt{7}}{2*6}=\frac{40-16\sqrt{7}}{12}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+16\sqrt{7}}{2*6}=\frac{40+16\sqrt{7}}{12}
$