(n-3)(n+4)=1

Solution for (n-3)(n+4)=1 equation:

(n-3)(n+4)=1

We move all terms to the left:

(n-3)(n+4)-(1)=0

We multiply parentheses ..

(+n^2+4n-3n-12)-1=0

We get rid of parentheses

n^2+4n-3n-12-1=0

We add all the numbers together, and all the variables

n^2+n-13=0

a = 1; b = 1; c = -13;
Δ = b2-4ac
Δ = 12-4·1·(-13)
Δ = 53
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{53}}{2*1}=\frac{-1-\sqrt{53}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{53}}{2*1}=\frac{-1+\sqrt{53}}{2}
$