(n+1)(n+2)=7n-1

Solution for (n+1)(n+2)=7n-1 equation:

(n+1)(n+2)=7n-1

We move all terms to the left:

(n+1)(n+2)-(7n-1)=0

We get rid of parentheses

(n+1)(n+2)-7n+1=0

We multiply parentheses ..

(+n^2+2n+n+2)-7n+1=0

We get rid of parentheses

n^2+2n+n-7n+2+1=0

We add all the numbers together, and all the variables

n^2-4n+3=0

a = 1; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2}{2*1}=\frac{2}{2}
=1
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2}{2*1}=\frac{6}{2}
=3
$