(n+4)(n-15)=0

Solution for (n+4)(n-15)=0 equation:

(n+4)(n-15)=0

We multiply parentheses ..

(+n^2-15n+4n-60)=0

We get rid of parentheses

n^2-15n+4n-60=0

We add all the numbers together, and all the variables

n^2-11n-60=0

a = 1; b = -11; c = -60;
Δ = b2-4ac
Δ = -112-4·1·(-60)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*1}=\frac{30}{2}
=15
$