(n+6)(n-2)=-9n+32

Solution for (n+6)(n-2)=-9n+32 equation:

(n+6)(n-2)=-9n+32

We move all terms to the left:

(n+6)(n-2)-(-9n+32)=0

We get rid of parentheses

(n+6)(n-2)+9n-32=0

We multiply parentheses ..

(+n^2-2n+6n-12)+9n-32=0

We get rid of parentheses

n^2-2n+6n+9n-12-32=0

We add all the numbers together, and all the variables

n^2+13n-44=0

a = 1; b = 13; c = -44;
Δ = b2-4ac
Δ = 132-4·1·(-44)
Δ = 345
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{-(13)-\sqrt{345}}{2*1}=\frac{-13-\sqrt{345}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{345}}{2*1}=\frac{-13+\sqrt{345}}{2}
$