(n+4)(n-6)=11

Solution for (n+4)(n-6)=11 equation:

(n+4)(n-6)=11

We move all terms to the left:

(n+4)(n-6)-(11)=0

We multiply parentheses ..

(+n^2-6n+4n-24)-11=0

We get rid of parentheses

n^2-6n+4n-24-11=0

We add all the numbers together, and all the variables

n^2-2n-35=0

a = 1; b = -2; c = -35;
Δ = b2-4ac
Δ = -22-4·1·(-35)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-12}{2*1}=\frac{-10}{2}
=-5
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+12}{2*1}=\frac{14}{2}
=7
$