x^2+12x+0=108

Solution for x^2+12x+0=108 equation:

x^2+12x+0=108

We move all terms to the left:

x^2+12x+0-(108)=0

We add all the numbers together, and all the variables

x^2+12x-108=0

a = 1; b = 12; c = -108;
Δ = b2-4ac
Δ = 122-4·1·(-108)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-24}{2*1}=\frac{-36}{2}
=-18
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+24}{2*1}=\frac{12}{2}
=6
$