(x+12)(x+4)=2

Solution for (x+12)(x+4)=2 equation:

(x+12)(x+4)=2

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2+16x+46=0

a = 1; b = 16; c = +46;
Δ = b2-4ac
Δ = 162-4·1·46
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{2}}{2*1}=\frac{-16-6\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{2}}{2*1}=\frac{-16+6\sqrt{2}}{2}
$