a^2+12a+4=0

Solution for a^2+12a+4=0 equation:

a^2+12a+4=0

a = 1; b = 12; c = +4;
Δ = b2-4ac
Δ = 122-4·1·4
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{2}}{2*1}=\frac{-12-8\sqrt{2}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{2}}{2*1}=\frac{-12+8\sqrt{2}}{2}
$