(a-3)(a+12)=0

Solution for (a-3)(a+12)=0 equation:

(a-3)(a+12)=0

We multiply parentheses ..

(+a^2+12a-3a-36)=0

We get rid of parentheses

a^2+12a-3a-36=0

We add all the numbers together, and all the variables

a^2+9a-36=0

a = 1; b = 9; c = -36;
Δ = b2-4ac
Δ = 92-4·1·(-36)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-15}{2*1}=\frac{-24}{2}
=-12
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+15}{2*1}=\frac{6}{2}
=3
$