(n-3)(n+10)=0

Solution for (n-3)(n+10)=0 equation:

(n-3)(n+10)=0

We multiply parentheses ..

(+n^2+10n-3n-30)=0

We get rid of parentheses

n^2+10n-3n-30=0

We add all the numbers together, and all the variables

n^2+7n-30=0

a = 1; b = 7; c = -30;
Δ = b2-4ac
Δ = 72-4·1·(-30)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*1}=\frac{-20}{2}
=-10
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*1}=\frac{6}{2}
=3
$