2(n-1)(2n-1)+(3n-1)(n+1)=21

Solution for 2(n-1)(2n-1)+(3n-1)(n+1)=21 equation:

2(n-1)(2n-1)+(3n-1)(n+1)=21

We move all terms to the left:

2(n-1)(2n-1)+(3n-1)(n+1)-(21)=0

We multiply parentheses ..

2(+2n^2-1n-2n+1)+(3n-1)(n+1)-21=0

We multiply parentheses

4n^2-2n-4n+(3n-1)(n+1)+2-21=0

We multiply parentheses ..

4n^2+(+3n^2+3n-1n-1)-2n-4n+2-21=0

We add all the numbers together, and all the variables

4n^2+(+3n^2+3n-1n-1)-6n-19=0

We get rid of parentheses

4n^2+3n^2+3n-1n-6n-1-19=0

We add all the numbers together, and all the variables

7n^2-4n-20=0

a = 7; b = -4; c = -20;
Δ = b2-4ac
Δ = -42-4·7·(-20)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-24}{2*7}=\frac{-20}{14}
=-1+3/7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+24}{2*7}=\frac{28}{14}
=2
$