-(6n-3)+7n(n+8)=1

Solution for -(6n-3)+7n(n+8)=1 equation:

-(6n-3)+7n(n+8)=1

We move all terms to the left:

-(6n-3)+7n(n+8)-(1)=0

We multiply parentheses

7n^2-(6n-3)+56n-1=0

We get rid of parentheses

7n^2-6n+56n+3-1=0

We add all the numbers together, and all the variables

7n^2+50n+2=0

a = 7; b = 50; c = +2;
Δ = b2-4ac
Δ = 502-4·7·2
Δ = 2444
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:

$\sqrt{\Delta}=\sqrt{2444}=\sqrt{4*611}=\sqrt{4}*\sqrt{611}=2\sqrt{611}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{611}}{2*7}=\frac{-50-2\sqrt{611}}{14}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{611}}{2*7}=\frac{-50+2\sqrt{611}}{14}
$