(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 add all the numbers together, and all the variables

(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=0

a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2}
=0
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2}
=2
$