4n(9-3n)=8n-1

Solution for 4n(9-3n)=8n-1 equation:

4n(9-3n)=8n-1

We move all terms to the left:

4n(9-3n)-(8n-1)=0

We add all the numbers together, and all the variables

4n(-3n+9)-(8n-1)=0

We multiply parentheses

-12n^2+36n-(8n-1)=0

We get rid of parentheses

-12n^2+36n-8n+1=0

We add all the numbers together, and all the variables

-12n^2+28n+1=0

a = -12; b = 28; c = +1;
Δ = b2-4ac
Δ = 282-4·(-12)·1
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-8\sqrt{13}}{2*-12}=\frac{-28-8\sqrt{13}}{-24}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+8\sqrt{13}}{2*-12}=\frac{-28+8\sqrt{13}}{-24}
$