-(x-12)(x-4)=0

Solution for -(x-12)(x-4)=0 equation:

-(x-12)(x-4)=0

We multiply parentheses ..

-(+x^2-4x-12x+48)=0

We get rid of parentheses

-x^2+4x+12x-48=0

We add all the numbers together, and all the variables

-1x^2+16x-48=0

a = -1; b = 16; c = -48;
Δ = b2-4ac
Δ = 162-4·(-1)·(-48)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*-1}=\frac{-24}{-2}
=+12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*-1}=\frac{-8}{-2}
=+4
$