4(n-3)n=8

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

4(n-3)n=8

We move all terms to the left:

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

We multiply parentheses

4n^2-12n-8=0

a = 4; b = -12; c = -8;
Δ = b2-4ac
Δ = -122-4·4·(-8)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{17}}{2*4}=\frac{12-4\sqrt{17}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{17}}{2*4}=\frac{12+4\sqrt{17}}{8}
$