x^2+12x+20=-4

Solution for x^2+12x+20=-4 equation:

x^2+12x+20=-4

We move all terms to the left:

x^2+12x+20-(-4)=0

We add all the numbers together, and all the variables

x^2+12x+24=0

a = 1; b = 12; c = +24;
Δ = b2-4ac
Δ = 122-4·1·24
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{3}}{2*1}=\frac{-12-4\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{3}}{2*1}=\frac{-12+4\sqrt{3}}{2}
$