(n+2)(n-1)=0

Solution for (n+2)(n-1)=0 equation:

(n+2)(n-1)=0

We multiply parentheses ..

(+n^2-1n+2n-2)=0

We get rid of parentheses

n^2-1n+2n-2=0

We add all the numbers together, and all the variables

n^2+n-2=0

a = 1; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·1·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3}{2*1}=\frac{-4}{2}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3}{2*1}=\frac{2}{2}
=1
$