(n+2)(n-7)=0

Solution for (n+2)(n-7)=0 equation:

(n+2)(n-7)=0

We multiply parentheses ..

(+n^2-7n+2n-14)=0

We get rid of parentheses

n^2-7n+2n-14=0

We add all the numbers together, and all the variables

n^2-5n-14=0

a = 1; b = -5; c = -14;
Δ = b2-4ac
Δ = -52-4·1·(-14)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*1}=\frac{-4}{2}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*1}=\frac{14}{2}
=7
$