(n-2)(3n-2)=4*16

Solution for (n-2)(3n-2)=4*16 equation:

(n-2)(3n-2)=4*16

We move all terms to the left:

(n-2)(3n-2)-(4*16)=0

We add all the numbers together, and all the variables

(n-2)(3n-2)-64=0

We multiply parentheses ..

(+3n^2-2n-6n+4)-64=0

We get rid of parentheses

3n^2-2n-6n+4-64=0

We add all the numbers together, and all the variables

3n^2-8n-60=0

a = 3; b = -8; c = -60;
Δ = b2-4ac
Δ = -82-4·3·(-60)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-28}{2*3}=\frac{-20}{6}
=-3+1/3
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+28}{2*3}=\frac{36}{6}
=6
$