(n+3)(n-5)=48

Solution for (n+3)(n-5)=48 equation:

(n+3)(n-5)=48

We move all terms to the left:

(n+3)(n-5)-(48)=0

We multiply parentheses ..

(+n^2-5n+3n-15)-48=0

We get rid of parentheses

n^2-5n+3n-15-48=0

We add all the numbers together, and all the variables

n^2-2n-63=0

a = 1; b = -2; c = -63;
Δ = b2-4ac
Δ = -22-4·1·(-63)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16}{2*1}=\frac{-14}{2}
=-7
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16}{2*1}=\frac{18}{2}
=9
$