2n-8(2n-7)+n(n-4)=66

Solution for 2n-8(2n-7)+n(n-4)=66 equation:

2n-8(2n-7)+n(n-4)=66

We move all terms to the left:

2n-8(2n-7)+n(n-4)-(66)=0

We multiply parentheses

n^2+2n-16n-4n+56-66=0

We add all the numbers together, and all the variables

n^2-18n-10=0

a = 1; b = -18; c = -10;
Δ = b2-4ac
Δ = -182-4·1·(-10)
Δ = 364
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:

$\sqrt{\Delta}=\sqrt{364}=\sqrt{4*91}=\sqrt{4}*\sqrt{91}=2\sqrt{91}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{91}}{2*1}=\frac{18-2\sqrt{91}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{91}}{2*1}=\frac{18+2\sqrt{91}}{2}
$