a^2+8a+16=12

Solution for a^2+8a+16=12 equation:

a^2+8a+16=12

We move all terms to the left:

a^2+8a+16-(12)=0

We add all the numbers together, and all the variables

a^2+8a+4=0

a = 1; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·1·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{3}}{2*1}=\frac{-8-4\sqrt{3}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{3}}{2*1}=\frac{-8+4\sqrt{3}}{2}
$