k^2+12k=-8

Solution for k^2+12k=-8 equation:

k^2+12k=-8

We move all terms to the left:

k^2+12k-(-8)=0

We add all the numbers together, and all the variables

k^2+12k+8=0

a = 1; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·1·8
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{7}}{2*1}=\frac{-12-4\sqrt{7}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{7}}{2*1}=\frac{-12+4\sqrt{7}}{2}
$