(n-5)(n+2)=0

Solution for (n-5)(n+2)=0 equation:

(n-5)(n+2)=0

We multiply parentheses ..

(+n^2+2n-5n-10)=0

We get rid of parentheses

n^2+2n-5n-10=0

We add all the numbers together, and all the variables

n^2-3n-10=0

a = 1; b = -3; c = -10;
Δ = b2-4ac
Δ = -32-4·1·(-10)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*1}=\frac{-4}{2}
=-2
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*1}=\frac{10}{2}
=5
$