(n-5)(n+3)=15

Solution for (n-5)(n+3)=15 equation:

(n-5)(n+3)=15

We move all terms to the left:

(n-5)(n+3)-(15)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

n^2-2n-30=0

a = 1; b = -2; c = -30;
Δ = b2-4ac
Δ = -22-4·1·(-30)
Δ = 124
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{124}=\sqrt{4*31}=\sqrt{4}*\sqrt{31}=2\sqrt{31}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{31}}{2*1}=\frac{2-2\sqrt{31}}{2}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{31}}{2*1}=\frac{2+2\sqrt{31}}{2}
$